MAT1035 Analytic Geometry

MAT1035 Analytic Geometry Lecture Notes R.A. Sabri Kaan Gürbüzer Dokuz Eylül University 2016

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Contents 1 Review of Trigonometry 5 2 Polar Coordinates 7 3 Vectors in R n 9 3.1 Located Vectors.............................................. 9 3.2 The Algebra of Vectors.......................................... 10 3.3 The Scalar Product............................................ 11 3.4 The Norm of a Vector........................................... 12 3.5 Projection of a Vector........................................... 13 3.6 The Direction Angles........................................... 14 3.7 Vector Product.............................................. 15 3.8 Mixed Product.............................................. 16 4 Straight Line 17 4.1 Direction of a Straight Line........................................ 17 4.2 Equations of a Straight Line....................................... 17 4.3 The Normal Form of a Straight Line in Plane.............................. 21 4.4 Intersection of Two Straight Lines in Space............................... 22 5 The Plane 25 5.1 The Plane Equation............................................ 25 5.2 The Angle between two Planes...................................... 27 5.3 Other Forms of the Equation of a Plane................................. 28 5.4 Distance from a Point to a Plane..................................... 29 5.5 Intersection of two Planes......................................... 31 5.6 Projecting Planes............................................. 32 5.7 Intersection of three Planes........................................ 32 5.8 Specialized Distance Formula...................................... 34 6 Transformation of Axes 37 6.1 Translation and Rotation of Axes..................................... 37 7 The Circle 39 7.1 The Equation of a Circle......................................... 39 7.2 Intersections Involving Circles...................................... 40 7.3 Systems of Circles............................................ 42 7.4 The Angle between two Circles...................................... 42 8 Ellipse, Hyperbola and Parabola 43 8.1 Ellipse................................................... 43 8.2 Hyperbola................................................. 44 8.3 Parabola.................................................. 46 3

4 CONTENTS 9 The Equations of Second Degree 49 10 Conic through Five Points and Family of Conics 51

Chapter 1 Review of Trigonometry P (cos θ, sin θ) hyp opp θ θ adj x 2 + y 2 = 1 P (cos θ, sin θ) Figure 1.1: The unit circle and trigonometric functions. cos θ = adj hyp, sin θ = opp hyp, tan θ = opp adj, cot θ = adj opp hyp hyp, sec θ =, csc θ = adj opp Let P (x, y) be a point at which the terminal side of θ intersects the unit circle x 2 + y 2 = 1. Then we have cos 2 θ + sin 2 θ = 1. Dividing each term of this fundamental identity by cos 2 θ gives the identity 1 + tan 2 θ = sec 2 θ. Dividing each term of the same identity by sin 2 θ gives 1 + cot 2 θ = csc 2 θ. The followings are the addition formulas 5

6 CHAPTER 1. REVIEW OF TRIGONOMETRY of the trigonometric functions. sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β sin α sin β sin 2θ = 2 sin θ cos θ cos 2θ = cos 2 θ sin 2 θ = 1 2 sin 2 θ = 2 cos 2 θ 1 cos 2 θ = 1 (1 + cos 2θ) 2 sin 2 θ = 1 (1 cos 2θ) 2 Angles frequently are measured in degrees with 360 in one complete revolution. In calsulus it is more convenient to measure angles in radians. The radian measure of an angle is the length of the arc it subtends in the unit circle when the vertex of the angle is at the center of the circle. The area of the unit disk with radius r is A = πr 2 and the circumference of the circle with radius r is C = 2πr. The circumference of the unit circle is 2π and its central angle is 360. It follows that 2πrad = 360 180 = πrad and 1 = 180. Consider an angle of θ radians at the center A π πr 2 = θ 2π give S = rθ and A = 1 2 r2 θ. The relationship between S of a circle of radius r. Then the proportions 2πr = the functions sin t and sin t is sin t = sin πt 180. Remember the properties of the trigonometric functions sin t + 2nπ = sin t and cos t + 2nπ = cos t for all n Z. To find the values of trigonometric functions of angles larger than π, we can use their periodicity and the following 2 identities. sin(π ± θ) = sin θ cos(π ± θ) = cos θ tan(π ± θ) = ± tan θ

Chapter 2 Polar Coordinates Let O be a fixed reference point called the pole and let L be a given ray called the polar axis beginning at 0. Now let us begin with a given xy-coordinate system and take the origin as the pole and the nonnegative x-axis as the polar axis. Definition 1. Given the pole O and the polar axis, the point P with polar coordinates r and θ, written as the ordered pair (r, θ) is loocated as follows: First find the terminal side of θ given in radians, where θ is measured counterclockwise (ifθ > 0) from the x-axis (the polar axis) as its initial side. If r 0, then P is on the terminal side of this angle at the distance r from the origin. If r < 0, then P lies on the ray opposite the terminal side at the distance r = r > 0 from the pole. P r θ θ r r > 0 r < 0 P Figure 2.1: Polar coordinates of a point with r > 0 and r < 0. The radial coordinate r can be described as the directed distance of P from the pole along the terminal side of θ. If r = 0, the polar coordinates (0, θ) represent the origin whatever the angular coordinate θ might be. 7

8 CHAPTER 2. POLAR COORDINATES Remark 1. Polar coordinates differ from rectangular coordinates in that any point has more than one representation in polar coordinates. Remark 2. The polar coordinates (r, θ + 2nπ) and ( r, θ + (2n + 1)π)), n Z, are the polar coordinates of the same point. The relation between polar coordinates and rectangular coordinates is given as follows: [ x y ] [ cos θ = r sin θ r = x 2 + y 2 tan θ = y x, x 0. ] The correct choice of θ : If x > 0, then (x, y) lies in either the first or fourth quadrant, so π 2 < θ < π 2 which is the range or the inverse tangent function. If x < 0 then (x, y) lies in the second or the third quadrant. So θ = arctan y x, x > 0 and θ = π + arctan y x, x < 0. Some curve have simpler equations in polar coordinates than in rectangular coordinates. The graph of an equation in the polar coordinate variables r and θ is the set of all those points P such that P has same pair of polar coordinates (r, θ) that satisfy the given equation {(r, θ) F (r, θ) = 0}. Example 1. The polar coordinate equation of the circle with center (0, 0) and radius a > 0 is r = a. If we start with the rectangular coordinate equation x 2 + y 2 = a 2 of this circle and transform it using x 2 + y 2 = r 2, we get r 2 = a 2. Example 2. Let us transform the equation r = 2 sin θ into rectangular coordinates. Now r 2 = 2r sin θ and so x 2 +y 2 = 2y x 2 + (y 1) 2 = 1 is a circle whose center is (0, 1) with radius is 1. More generally, the graphs of the equations r = 2a sin θ and r = 2a cos θ are circles of radius a centered, respectively, at (0, a) and (a, 0). Example 3. We can transform the rectangular equation ax + by = c of a straight line into ar cos θ + br sin θ = c. Let us take a = 1, b = 0. The polar equation of the vertical line x = c is r = c sec θ where π 2 < θ < π 2. Example 4. Sketch the graph of the polar equation r = 2 + 2 sin θ.

Chapter 3 Vectors in R n The quantities, such as distance, area, volume, temprature, time densty, etc., posses only magnitude. These can be represented by real numbers and are called scalars. On the other hand, the quantities, such as force, velocity, acceleration, etc., posses both magnitude and direction. These quantities can be represented by arrows and are called vectors. Moreover, the elements of a vector space are called the vectors. 3.1 Located Vectors Let A and B be the point in the space R 3. We can denote a vector in R 3 by an ordered triple. Hence, A = (a 1, a 2, a 3 ), B = (b 1, b 2, b 3 ). We define a located vector to be an ordered pair of points, for example, (A, B) and denote it by AB. We visualize this an arrow from A to B. We call the point A the beginning point and the point B the end point of the located vector AB. z A AB B y x Figure 3.1: Located vector from the point A to the point B in the three dimensional space. Now, let us denote the located vector AB by the order triple (b 1 a 1, b 2 a 2, b 3 a 3 ). But this correspondence is not one to one. An other located vector, such as CD, can also be identified by the same ordered triple. It is clear that, the set of located vectors forms an equivalence class with the equivalence relation having the following properties: i. reflexive, that is, AB AB ii. symmetric, that is, AB CD CD AB. iii. transitive, that is, AB CD CD EF AB EF. Each equivalence class called a vector. Therefore AB = B A = O(B A) and b i a i are called the coordinates of AB. 9

10 CHAPTER 3. VECTORS IN R N Question 1. Determine the beginning point of the vector (2, 6, 0) whose end point is (1, 1, 3). Question 2. For what values of a, b and c are the vectors (2a b, a 2b, 6) and ( 2, 2, a + b 2c) equal? 3.2 The Algebra of Vectors Two vectors a = (a 1, a 2, a 3 ) and b = (b 1, b 2, b 3 ) are equivalent if and only if a 1 = b 1, a 2 Geometrically, a and b have the same direction and their magnitudes (or lengths) are equal. = b 2, a 3 = b 3. The addition of vectors is defined by coordinate-wise, that is, a + b = (a1 + b 1, a 2 + b 2, a 3 + b 3 ) and addition can geometrically be considered by the following triangle or parallelogram laws. a a b a a + b a + b b b Triangle Law Prallelogram Law Figure 3.2: Addition of vectors Since the vectors are the elements of a vector space, we can also define b. Geometrically, b has the opposite direction of b but equal size. Two vectors a and b in R n are parallel if and only if a = k b, where k R. Question 3. Compute u + v, u v, 2 u 3 v, 3 u + 1 2 v and sketch them if u = (2, 1, 4) and v = (1, 2, 3). Question 4. Show that AB + BC + CA = 0 for a triangle ABC. Question 5. Show that AD = 1 2 ( AC + AB) for the median vector AD of a triangle ABC.

3.3. THE SCALAR PRODUCT 11 Question 6. Let the points A, B and C be collinear and let the point D do not lie on the line containing the points A, B and C. If AC = 3 AB and DC = kad + pdb then k p =? Question 7. Let G be a barycenter of a triangle ABC. Show that a) GA + GB + GC = 0 b) for any point K, KA + KB + KC = 3 KG. Question 8. Let ABC be a triangle, and D and E be the midpoints of AB and AC. Then show that DE = 1 BC 2. Question 9. If the vectors u = (1, 2 m, n m) and v = (3, 3m, 8m) are parallel, then find the values of m and n. 3.3 The Scalar Product Let a, b R n. We define their scalar (or dot or inner) product a b to be a bilinear map R n R n R such that ( a, b ) a b := a 1 b 1 + a 2 b 2 + + a n b n = n a i b i (3.1) i=1 0. We define two vectors a and b to be orthogonal (or perpendicular) denoted by a b if and only if a b = Question 10. Let V be a vector space and <, > an inner product on V. a) Show that < 0, u >= 0 for all u in V. b) Show that < u, v >= 0 for all v in V, then u = 0.

12 CHAPTER 3. VECTORS IN R N Question 11. Let <, > be standard inner product on R 2. a) Let α = (1, 2) and β = ( 1, 1). Find a vector γ such that < α, γ >= 1 and < β, γ >= 3. b) Show that α =< α, e 1 > e 1 + < α, e 2 > e 2 for any α in R 2. (Here, the vectors e 1 and e 2 are standard basis vectors of R 2.) 3.4 The Norm of a Vector The norm of length of a vector a is denoted by a and is defined by the number a a, that is, a := a a = a1 a 1 + + a n a n = a 2 1 + + a2 n = n a 2 i (3.2) i=1 Example 5. Let the vectors u = (3, 1), v = (2, 3), w = (4, 5) be given. ( w u ) v = 4 2 + ( 5) 2 (3 2 + ( 1) 3) = 3 41 v w = (2 4, 3 ( 5)) = ( 2) 2 + 8 2 = 68 We shall say that a vector e is a unit vector if e = 1. Also we can normalize a vector a dividing it by its norm, that is, a e = a. Question 12. Find the norm of α = (3, 4) with respect to a) the usual inner product. b) the inner product given as < x, y >= x 1 y 1 x 1 y 2 x 2 y 1 + 3x 2 y 2. Question 13. Find the vector AB and CD, and their lengths, and also sketch them in R 3 if A( 2, 1, 0) and B( 1, 1, 7) and C(5, 2, 3) and D(2, 1, 0). Theorem 1. For u, v R n, u v u v.

3.5. PROJECTION OF A VECTOR 13 Proof. Let f : R R be a function such that f(t) = u + t v 2 = u u + 2 u t v + t 2 ( v v ). Clearly, the function f is a second degree polynomial of t and since we consider a square of a reel number, f has at most one root. So 4( u v ) 2 4( u u )( v v ) 0 ( u v ) 2 ( u u )( v v ) u v u v Theorem 2. For u, v R n, u + v u + v. Proof. By using Cauchy-Schwartz inequality, we get u + v 2 = u u + 2 u v + v v u u + 2 u v + v v = ( u + v ) 2 Since both sides of inequality are square of positive numbers, we have u + v u + v. 3.5 Projection of a Vector Let a and b be two vectors in the same space and b 0. We wish to find the projection of a along b, that is, the vector p in the following figure. a p a p b Figure 3.3: Projection of a vector It is clear that p = c b, for some c R. On the other hand, a p b. Hence we have the following computations. ( a c b ) b = 0 a b c b b = 0 a b = c b b c = a b b 2 (3.3) The vector p := c b = ( a b ) b is called the projection of a along b and the scalar c is called the b 2 component of a along b. Our construction gives us a geometric interpretation for the scalar product. Let θ be the acute angle between a and b. The cosine of θ is computed as follows: cos θ = c b a = c b a = ( a b ) b a b b 2 a = a b (3.4)

14 CHAPTER 3. VECTORS IN R N a c b a θ c b b Figure 3.4: Angle between two vectors Example 6. Let u = (3, 1) and v = ( 1, 2). Then cos α = 3 ( 1) + 1 ( 2) 3 2 + 1 2 ( 1) 2 + ( 2) = 5 = 1 2 50 2 Question 14. Find the projection of the vector u = (1, 2, 2) onto v = ( 3, 4, 5)). Question 15. Find the angle between the vectors u and v if u = (1, 1, 2) and v = ( 1, 1, 1). Question 16. Let ABC be a triangle. Prove the cosine theorem. 3.6 The Direction Angles The unit vectors e i = (0, 0,..., 0, 1, 0,..., 0, 0), where 1 sits in the i-th coordinate, are called the natural base vectors. The natural base vectors are mutually perpendicular. Consider the vector a making the angles α 1,..., α n with the natural base vectors e 1,..., e n, respectively. These angles are called the direction angles of a and their cosines are called the direction cosines of a. Since each vector is a linear combination of the natural base vectors, a e i = a i. So the cosine of α i is computed by cos α i = a i a. The direction cosines are related by the equation n i=1 cos2 α i = 1. For 1 i n, the coordinate a e i = a i is called the direction number of a. Any set proportional to the coordinates are also called the direction numbers of a, such as ka 1,..., ka n, where k R.

3.7. VECTOR PRODUCT 15 3.7 Vector Product Let a and b be two vectors in R 3. The vector product or cross product of a and b is a antisymmetric bilinear map R 3 R 3 R 3 defined by a b = (a2 b 3 a 3 b 2, a 3 b 1 a 1 b 3, a 1 b 2 a 2 b 1 ) (3.5) The cross product of the natural base vectors in R 3 is given as follows: e1 e 2 = e 3 e2 e 1 = e 3 e2 e 3 = e 1 e3 e 2 = e 1 e3 e 1 = e 2 e1 e 3 = e 2 We can simply get the same result by the Sarrus Rule using the following determinant. e1 e2 e3 a b = a 1 a 2 a 3 b 1 b 2 b 3 (3.6) Question 17. Compute the followings: x y if x = (1, 2, 1) and y = (1, 1, 1). u v if u = (1, 5, 8) and v = ( 2, 4, 1). Question 18. If u v = u ω then can we necessarily say that v = ω? Let a, b, c, d R 3 and k R. The vector product satisfies the following properties. 1. 2. a b = b a. a ( b + c ) = ( a b ) + ( a c ) and ( a + b ) c = ( a c ) + ( b c ). 3. (k a ) b = k( a b ) = a (k b ). 4. ( a b ) c = ( a c ) b ( b c ) a. 5. a b is perpendicular to both a and b. 6. ( a b ) ( c d ) = ( a c )( b d ) ( a d )( b c ). 7. a // b in R 3 if and only if a b = 0. 8. a b = a b sin θ, where θ is the acute angle between the vectors a and b. Question 19. Let u, v and ω be arbitrary vectors in 3-space. Is there any difference between ( u v) ω and u ( v ω)? The geometric interpretation of the last property may be given as the area of the parallelogram formed with the vectors a and b as two adjacent sides.

16 CHAPTER 3. VECTORS IN R N a θ h b Figure 3.5: The parallelogram formed by two vectors 3.8 Mixed Product The combination of a vector product and a scalar product is called mixed product (or triple product). The mixed product of the vectors a, b, c R 3 is defined by ( a, b, c ) := a ( b a 1 a 2 a 3 c ) = b 1 b 2 b 3 (3.7) c 1 c 2 c 3 Since interchanging the adjacent rows of a determinant changes the sign of the determinant, we have the following equalities. ( a, b, c ) = ( b, a, c ) = ( b, c, a ) = ( c, b, a ) = ( c, a, b ) = ( a, c, b ) The geometric interpretation of triple product may be given as the volume of the parallelepiped formed with the vectors a, b and c as adjacent edges. a h θ c b Figure 3.6: The Parallelepiped formed with the vectors in three dimensional space.

Chapter 4 Straight Line 4.1 Direction of a Straight Line The direction angles of a straight line are the direction angles of a vector lying on that line. If α 1,..., α n are direction angles of a line then for some λ R different form zero, we have a 1 = λ cos α 1,..., a n = λ cos α n. The numbers a 1,..., a n are called a set of direction numbers of the line and the vector a = (a 1,..., a n ) is called the direction vector of the line. It is clear that the component of any vector are a set of direction numbers of any line on which the vector lies. Conversely, if a 1,..., a n are direction numbers of a line then the vectors (a 1,..., a n ) and ( a 1,..., a n ) are oppositely directed vectors on that line. Theorem 3. Let a 1,..., a n and b 1,..., b n be the the direction numbers of two lines l 1 and l 2, respectively. i. The lines l 1 and l 2 are perpendicular if and only if n a i b i = 0. ii. The lines l 1 and l 2 are parallel if and only if a 1 = a 2 = = a n. b 1 b 2 b n n i=1 The angle θ between two lines l 1 and l 2 is given by cos θ = a ib i n n. i=1 a2 i i=1 b2 i i=1 4.2 Equations of a Straight Line Let a 1,..., a n be a set of direction numbers of a line containing the point P 0 (p 1,..., p n ), then a = (a 1,..., a n ) is a vector parallel to the line. Our discussion is to find the locus of the points lying on the straight line. We take an arbitrary point to determine all the points lying on the straight line. Let the arbitrary point be P (x 1,..., x n ). According to the Figure 4.1, we have the equation OP = OP0 + t a, where the parameter t R. Hence we have the vector equation (vector form) of the straight line. which is also written as follows. (x 1,..., x n ) = ((p 1,..., p n ) + t(a 1,..., a n ) (4.1) x 1 = p 1 + ta 1. (4.2) x n = p n + ta n 17

18 CHAPTER 4. STRAIGHT LINE x n a P 0 (p 1,..., p n ) l P (x 1,..., x n ) O x 2 x Figure 4.1: Straight Line in n-space The equations 4.2 are called the parametric equations of the line in terms of a point P 0 (p 1,..., p n ) and a set of direction numbers a 1,..., a n. If none of the quantities a 1,..., a n vanishes we may solve each of the equations in parametric equation for t and obtain x 1 p 1 a 1 = = x n p n a n (= t). (4.3) The equation 4.3 is called the symmetric equation of the line passing through the point P 0 (p 1,..., p n ) with direction a = (a1,..., a n ). Question 20. Under which condition the points A = (x 1, y 1, z 1 ), B = (x 2, y 2, z 2 ) and C = (x 3, y 3, z 3 ) are collinear? Question 21. Let P 1 and P 2 be two distinct points in 3-space. a) Show that P = tp 1 + (1 t) P 2, t R is a vector equation of the line passing through P 1 and P 2. b) Show that the line segment joining P 1 to P 2 is given by [P 1, P 2 ] = tp 1 + (1 t) P 2, t [0, 1]. Question 22. Write the equations of the line whose parametric equations are x = 5 2t, y = 3 + t, z = 2 3t in symmetric form. Question 23. Write the equations of the line through the points (7, 1, 2) and (3, 2, 4). Question 24. Write the equations of the line parallel to x 1 2 = y + 4 = z 3 and through (4, 2, 3).

4.2. EQUATIONS OF A STRAIGHT LINE 19 as In 2-space we can express the equations of a line passing through the point P 0 (p 1, p 2 ) with direction numbers a, b We can express this equation as x p 1 a = y p 2. (4.4) b letting m = b a, y p 2 = m(x 1 + p 1 ) and letting n = p 2 mp 1, y = b a (x p 1) + p 2 (4.5) y = mx + n. (4.6) The ratio m = tan α = b a is called the slope of the straight line. The equation 4.5 is called the point-slope form of the straight line and the equation 4.6 is called the slope-intercept form of the straight line. In equation 4.6, the y-intercept is (0, n). On the other hand, given two points define a direction, hence a straight line passing through these points. For instance in R 2 let the points P 0 (p 1, p 2 ) and Q 0 (q 1, q 2 ) be given. Then we have a straight line passing through P 0 and Q 0 with direction d = P 0 Q 0 and the equation OP = OP0 + t P 0 Q 0. One can also use the equality of the slope tan α = q 2 p 2 q 1 p 1 = y p 2 x p 1. (4.7) The equation 4.7 is called the two point form of a line. The equation of the straight line whose x and y-intercepts are respectively (a, 0) and (0, b) is x a + y b = 1. (4.8) The equation 4.8 is called the intercept form of the line. Every equation of the first degree in x and y, that is, a straight line in R 2, may be reduced to the form ax + by + c = 0, (4.9) where a, b, c are arbitrary constants. The equation 4.9 is called general form of a line in R 2. Theorem 4. Two nonvertical lines l 1 and l 2 in R 2 are parallel if and only if their slopes are equal. Two nonvertical lines are perpendicular if and only if the slpoe of one is the negative of the reciprocal of the slope of the other, that is, two lines with slopes m 1 and m 2 are perpendicular to each other if and only if m 2 = 1 m 1. Question 25. Find the coordinates of midpoint M(x 1, x 2 ) of the line segment joining the points P 1 (3, 7) and P 2 ( 2, 3). Question 26. Find the coordinates of the point P (x, y) which divides the line segment joining the points P 1 ( 2, 5) and P 2 (4, 1), in the ratio at 6 5 and 2, respectively.

20 CHAPTER 4. STRAIGHT LINE Question 27. Find the slope and direction cosines of the line which is perpendicular to the line joining the points P 1 (2, 4) and P 2 ( 2, 1). Question 28. Find the angle between the directed lines joining P 1 (1, 3), P 2 ( 4, 3), and P 3 (2, 0) and P 4 ( 5, 6) by slopes and direction cosines. Question 29. Given the triangle whose vertices are A( 5, 6), B( 1, 4) and C(3, 2) derive the equations of the three medians and solve algebraically for their point of intersection. Question 30. Find the intercepts of the line perpendicular to 2x + 3y 7 = 0 and passing through the point (1, 6). Question 31. Find the equations of the lines through (1, 6) if the product of the intercepts for each line is 1. Question 32. What are the direction cosines of a line perpendicular to x 5y + 3 = 0. Question 33. Find the equations of all the lines with slope m = 3 4 such that each line will occur an area 24 unit2 with the coordinate axes. Question 34. Determine the parameter k such that a) the line 3kx + 5y + k 2 = 0 passes through A( 1, 4). b) the line 4x ky 7 has the slope 3. c) the line whose equation 2x + ky + 3 = 0 shall make an angle 45 o with the line 2x + 5y 17 = 0.

4.3. THE NORMAL FORM OF A STRAIGHT LINE IN PLANE 21 4.3 The Normal Form of a Straight Line in Plane Let l be a straight line which does not pass through the origin. If the length of the perpendicular drawn from the origin to the line l is p and the angle that the perpendicular makes with the x-axis is ω then we find the equation of the line l. y y 1 A(x 1, y 1 ) p Q(x, y) ω O x 1 x Figure 4.2: Normal form of a straight line Let A(x 1, y 1 ) be the intersection of the perpendicular line segment and the line l. Our aim is to define the arbitrary 1 point Q(x, y) on the line. Form the Figure, x 1 = p cos ω, y 1 = p sin ω and the slope of l is. Then by the tan ω equation 4.5, y y 1 = 4.10. 1 tan ω (x x 1) or y p sin ω = 1 (x p cos ω). This equation reduces to the equation tan ω y sin ω + x cos ω = p (4.10) The equation 4.10 with p > 0 is called the normal form of a line not passing through the origin. If ax + by + c = and x cos ω + y sin ω p = 0 are the general and normal form of the same straight line l then the coefficients of the two equations are proportional. Hence cos ω a = sin ω b = p c = k, 1 where k is the constant ratio. Then we have cos ω = ka, sin ω = kb and k = ± a 2 + b. Hence 2 cos ω = ±a a 2 + b 2 sin ω = Then the normal form of ax + by + c = 0 is ±b a 2 + b 2 p = c a 2 + b 2. a ± a 2 + b x + b 2 ± a 2 + b + c 2 ± = 0. (4.11) a 2 + b2 Since p > 0, the sign before the square root is chosen the opposite sign of c. To find the perpendicular distance d from the line l to the point x 1, y 1, draw the line L 1 through the point (x 1, y 1 ) which is parallel to l. If the normal equation of l is x cos ω + y sin ω = p then the normal equation of L 1 is x cos ω + y sin ω = p + d. Since (x 1, y 1 ) L 1, the coordinates (x 1, y 1 ) satisfy the equation of L 1, x 1 cos ω + y 1 sin ω = p. Hence the distance d is obtained by d = x 1 cos ω + y 1 sin ω p. (4.12) If (x 1, y 1 ) and the origin are on opposite sides of the line l then the distance d is positive and if they are on the same side of the line l then d is negative.

22 CHAPTER 4. STRAIGHT LINE y L 1 l O p d (x 1, y 1 ) x Figure 4.3: Distance from a point to a line Question 35. Find the perpendicular distances from origin to the lines a) x 3y + 6 = 0 b) 15x 8y 25=0 Question 36. Find the equation of the line through the point of intersection of the lines x 3y+1 = 0 and 2x+5y 9 = 0 and whose distance from the origin is 5. Question 37. Calculate the distance d from the line 5x 12y 3 = 0 to point (2, 3). Are the point (2, 3) and the origin on the same side? Question 38. Determine the equations of the bisectors of the angles between the lines l 1 : l 2 : 12x 5y + 5 = 0. 3x + 4y 2 = 0 and 4.4 Intersection of Two Straight Lines in Space Two straight line in space may be parallel or not, which is easily determined by their direction numbers. If they are not parallel then they may still have no point of intersection, in which case they are called skew lines. We shall illustrate the method of intersection of straight lines by using the parametric form. Let us discuss the intersection of the two lines l 1 : x x 1 = y y 1 = z z 1 (= t) and l 2 : x x 2 = y y 2 = z z 2 (= u). The a 1 b 1 c 1 a 2 b 2 c 2

4.4. INTERSECTION OF TWO STRAIGHT LINES IN SPACE 23 l 1 l 2 l 1 = l 2 l 1 l 2 l 1 l 2 d1 d2 = d 1 d2 = d 1 d1 d2 d1 d2 Figure 4.4: Position of two lines in 3-space parametric equations are x = x 1 + ta 1 x = x 2 + ua 2 y = y 1 + tb 1 y = y 2 + ub 2 z = z 1 + tc 1 z = z 2 + uc 2 We have to discuss whether they have common point(s) or not. Hence x = x 1 + ta 1 = x 2 + ua 2 a 1 t + a 2 u = x 1 x 2 y = y 1 + tb 1 = y 2 + ub 2 b 1 t + b 2 u = y 1 y 2 z = z 1 + tc 1 = z 2 + uc 2 c 1 t + c 2 u = z 1 z 2 a 1 a 2 b 1 b 2 and b = c 1 c 2 There are 3 equations and 2 unknowns (u, t). Let A = x 1 x 2 y 1 y 2 z 1 z 2 Remark 3. The system is consistent if Rank(A) = Rank(A : b), that is, we have solution for Rank(A) = Rank(A : b). Here we have two cases If Rank(A) = Rank(A : b) = 2 then the system has a unique solution and hence the lines intersect at a unique point. If Rank(A) = Rank(A : b) = 1 we have infinite solutions, hence the lines intersect at infinitely many points, that is, the lines are coincide. If Rank(A) < Rank(A : b) then the system has no solution and hence the straight lines have no point in common, which means that they are parallel or skew. Question 39. Discuss the intersection of the lines l 1 and l 2 if a) l 1 : x + 1 = y 1 = z 3 2 and l 2 : 2x + 2 = 6 2y = z + 1. b) l 1 : x + 1 = y 1 = z 3 2 and l 2 : 2x 4 = 2y 2 = z 5. c) l 1 : x + 1 = y 1 = z 3 2 and l 2 : x 3 2 = y 4 = z 1 2. Question 40. x 1 Given l = {(x, y, z) : 2 = y+2 1 = z 3 }, find a line l through (1, 2, 3) such that l and l are skew.

24 CHAPTER 4. STRAIGHT LINE

Chapter 5 The Plane 5.1 The Plane Equation Let P 0 (x 0, y 0, z 0 ) be a fixed point and let P (x, y, z) be an arbitrary point in the plane. Let a, b, c be a set of direction numbers of any line perpendicular to the plane. Let n be the vector n = (a, b, c) which is direction vector of the line perpendicular to the plane. Since P 0 P lies in the plane, n is perpendicular to P 0 P. Hence n P0 P = n ( P P 0 ) = 0. (5.1) n P 0 P 0 P P z x O y Figure 5.1: Vector form of a plane The equation 5.1 is called the vector form of the equation of the plane which contains the point P 0 and is perpendicular to the direction n. The vector n is called the normal to the plane. If the vectors n, P and P 0 are expressed in terms of their coordinates then we obtain the nonvector forms of the plane given in the following equations n P0 P = 0 n ( P P 0 ) = 0 (a, b, c) (x x 0, y y 0, z z 0 ) = 0 a(x x 0 ) + b(y y 0 ) + c(z z 0 ) = 0 (5.2) Let d = (ax 0 + by 0 + cz 0 ) ax + by + cz + d = 0 (5.3) 25

26 CHAPTER 5. THE PLANE In each of the equations 5.2 and 5.3, the coefficients a, b and c are a set of direction numbers of a normal to the plane. The equation 5.2 is called the standard form and the equation 5.3 is called the general form of the equation of the plane. Theorem 5. Any plane may be represented by a linear equation and conversely any linear equation in R 3 represents a plane. Proof. The direct statement has been proved by driving the equation 5.3. To prove the converse, suppose that we have a linear equation ax + by + cz + d = 0. We may assume that c 0 and then write ax + by + c(z d c ) = 0 or a(x 0) + b(y 0) + c(z d c ) = 0. It follows that the vector through P (x, y, z) and P 0(0, 0, d ) is perpendicular to c the fixed direction n = (a, b, c). This is just a plane equation passing through P 0 with normal n. Question 41. Under which condition, four points A = (x 1, y 1, z 1 ), B = (x 2, y 2, z 2 ), C = (x 3, y 3, z 3 ) and D = (x 4, y 4, z 4 ) are coplanar, i.e. lie on a same plane? Question 42. For which value of λ, the points A(2, 3, λ), B(1, 0, 2), C(2, 1, 2) and D(1, 1, 2) are coplanar? Question 43. Find the equation of the plane perpendicular to the line joining (1, 3, 5) and (4, 3, 2) at the midpoint of these two points. Question 44. Write equation of the plane passing through (1, 1, 2) and (3, 5, 4) which is perpendicular to the xy-plane. Question 45. Find the equation of the plane perpendicular to the plane x y + 2z 5 = 0, parallel to the line whose direction cosines are 1 5, 2 5, 2 5 5 and passing through (1, 4, 1). Question 46. Find the equation of the plane parallel to the 3x 2y + 6z + 5 = 0 and passing through (1, 4, 1).

5.2. THE ANGLE BETWEEN TWO PLANES 27 Question 47. Find the equation of the plane parallel to the xz-plane and through (1, 4, 6). Question 48. Find an equation of the plane through (1, 2, 3) and perpendicular to the line l = {(x, y, z) : x 1 z 3 }. 2 = y+2 1 = Question 49. Find an equation of the plane through (1, 2, 3) and l = {(x, y, z) : x 1 2 = y+2 1 = z 3 }. Question 50. Find an equation of the plane determined by the lines l 1 = {(x, y, z) : l 2 = {(t 2, 2t 1, 3t + 3) : t R}. x + 1 = y + 1 = z} and 5.2 The Angle between two Planes By definition, the angle between two planes is the angle between their normals. For the given planes P 1 : a 1 x + b 1 y + c 1 z + d 1 = 0 P 2 : a 2 x + b 2 y + c 2 z + d 2 = 0 the direction numbers to the normals are n 1 = (a 1, b 1, c 1 ) and n 2 = (a 2, b 2, c 2 ), respectively. Then we have n 1 n 2 cos α = n 1 n 2 = a 1 a 2 + b 1 b 2 + c 1 c 2 a 2 1 + b 2 1 + c2 1 a 2 2 + b 2 2 +. (5.4) c2 2 n 2 n 1 α P 1 P 2 Figure 5.2: Angle between two planes If α = π in 5.4 then the planes are perpendicular to each other. If α = 0 or α = π then the normals to the plane 2 are parallel to each other and so the planes are parallel.

28 CHAPTER 5. THE PLANE Question 51. Determine the angle between the planes x + y z + 7 = 0 and 3x y 6 = 0. Question 52. Find the cosine of the angle between the two planes with equations a) P 1 : x + y + z 1 = 0 and P 2 : 2x y + 2z + 4 = 0, b) P 3 : x + 2y + z 1 = 0 and P 4 : 4x 2y z 3 = 0, c) P 5 : x 6y + 2z 1 = 0 and P 6 : x 2y + 2z 3 = 0. 5.3 Other Forms of the Equation of a Plane If a plane intersects a coordinate plane in a straight line, that line is called a trace of that plane. If a plane intersects a coordinate axis in a point, that point is called an intercept of the plane. The intercepts may be found from the equations of the traces and the general form of the equation of the plane. Suppose a, b, c and d are all different from zero in the equation ax+by +cz +d = 0. Calling x, y and z-intercepts A, B and C, respectively, we find that A = d a, B = d b and C = d. Dividing each coefficient in the general form ax + by + cz + d = 0 by d and taking the reciprocals c x we get ( d a ) + y ( d b ) + z ( d + 1 = 0. Hence the general form may be written in the form c ) x A + y B + z C = 1. (5.5) The equation 5.5 is called the intercept form of the equation of the plane. Let us write the vector form of the equation of a plane as n P = t, (5.6) where n is the normal vector of the plane and t = n P 0. Dividing both sides by ± n and using the sign of t we n have ± n t P = ± n. If we set n N = ± n and p = t ± n, it follows that p > 0 and N is a unit vector. The equation 5.6 may be written as N P = p, p > 0. (5.7) The equation 5.7 is the vector normal form of the equation of a plane. Since N is a unit vector, it may be written in terms of its direction angles as N = (cos α, cos β, cos γ). It is easily seen that the vector R 0 = p N satisfies the equation 5.7. The corresponding point p N = (p cos α, p cos β, p cos γ) lies in the plane, p N is perpendicular to the plane and p N = p. In nonvector form, the equation 5.7 becomes (cos α, cos β, cos γ) (x, y, z) = p, p > 0 x cos α + y cos β + z cos γ = p, p > 0. (5.8) The equation 5.8 is called the normal form of the equation of a plane, where α, β and γ are the direction angles of the normal vector N and p is the directed distance from origin to the plane. One can convert the general form ax + by + cz + d = 0 of the equation of a plane to the normal form by dividing the equation by a 2 + b 2 + c 2 and using the opposite sign of d.

5.4. DISTANCE FROM A POINT TO A PLANE 29 z R 0 = p N γ α O β y x Figure 5.3: Vector normal form of a plane Question 53. Reduce 3x + 2y z + 5 = 0 to normal form. Question 54. Find the equation of the plane parallel to the xy-plane and 2 units from it. Question 55. Reduce x 5y + 2z 3 = 0 to intercept form and write the coordinates of the intercepts. Question 56. Find the equation of the plane with intercepts 1, 2, 4. 5.4 Distance from a Point to a Plane Similar to the normal form of the straight line in R 2 one can simply obtain the distance from a point to a plane by normal form of its equation. We want to find the distance l of the point P 0 to the plane ax + by + cz + d = 0. Let P 1 be the foot of the perpendicular drawn from P 0 to the plane then OP 0 = OP 1 + P 1 P 0. Since n // P 1 P 0, taking the inner product of both sides by n, we get

30 CHAPTER 5. THE PLANE n = (a, b, c) ax + by + cz + d = 0 z OP 0 OP 1 P 0 (x 0, y 0, z 0 ) P 1 P 0 P 1 (x 1, y 1, z 1 ) O y x Figure 5.4: Distance from a point to a plane. n OP0 = n OP 1 + n P 1 P 0 (a, b, c) (x 0, y 0, z 0 ) = (a, b, c) (x 1, y 1, z 1 ) + n P 1 P 0 cos α ax 0 + by 0 + cz 0 = ax 1 + by 1 + cz 1 ± n P 1 P 0 ax 0 + by 0 + cz 0 = d ± n.l ax 0 + by 0 + cz 0 + d = ± n.l l = ± ax 0 + by 0 + cz 0 + d = ± ax 0 + by 0 + cz 0 + d. n a 2 + b 2 + c 2 Two points lie on the same side of the plane ax + by + cz + d = 0 if the left member of ax + by + cz + d = 0 has the same sign for both points and they are opposite sides if the left member has opposite signs. Question 57. Find the distance from the plane x y + z + 2 = 0 to the point (0,-2,3). Question 58. Find the distance from P 0 = (3, 3, 2) to the plane P = {(x, y, z) : x + 2y 2z + 8 = 0}. Question 59. Determine the locations (i.e. whether they are on the same side with the origin or not) of the points A(4, 5, 4), B( 5, 8, 6) and C(3, 25, 4) with respect to the plane P : 2x 3y + z + 7 = 0.

5.5. INTERSECTION OF TWO PLANES 31 5.5 Intersection of two Planes Let two planes a 1 x + b 1 y + c 1 z + d 1 = 0 and a 2 x + b 2 y + c 2 z + d 2 = 0 be given. These two planes are parallel if and only if a 1 = b 1 = c 1 and they are coinciding if and only if a 1 = b 1 = c 1 = d 1. If these two planes are a 2 b 2 c 2 a 2 b 2 c 2 d 2 not parallel, they intersect in a straight line which consists of those points satisfying both equations of the planes. We have three methods for finding the intersection line of these planes. 1. First Method: Find any two points from the intersection of the planes and then find the equation of the intersection line. 2. Second Method: Use the normals to the planes to find the direction of the straight line. Intersection line is perpendicular to the normals of both planes ( n 1, n 2 ). Hence the direction of the intersection line is n 1 n 2. After finding a common point satisfying both of the equations of planes, one can easily find the equation of the intersection line. 3. Third Method: We eliminate one variable from the equations and then eliminate an other one. Hence we may write the symmetric equations of the straight line. Question 60. Determine l P if l and P are given by l = {(t + 1, t 2, 3t + 1) : t R} and P = {(x, y, z) : x y + 2z + 1 = 0}. Question 61. Determine l P if l and P have equations: a) l = {( t 1, t + 2, 2t 1) : t R} and P = {(x, y, z) : x 2y + 3z + 4 = 0}. b) l = {(x, y, z) : x = y+2 2 = z 3 } and P = {(x, y, z) : x + 2y + 3z 4 = 0}. c) l = {(x, y, z) : x+6 3 = y 2 = z 1} and P = {(x, y, z) : x 2y + z 4 = 0}. Question 62. Find the vector formulation for the line of intersection of the planes P 1 : x + 2y + 3z + 4 = 0 and P 2 : x 2y + 2z 1 = 0. Question 63. Find a vector equation of the line of intersection of the following planes: a) P 1 : x + y + z 1 = 0 and P 2 : 2x y + 2z + 4 = 0, b) P 3 : x + 2y + z 1 = 0 and P 4 : 4x 2y z 3 = 0, c) P 5 : x 6y + 2z 1 = 0 and P 6 : x 2y + 2z 3 = 0.

32 CHAPTER 5. THE PLANE Question 64. Discuss the intersection of given planes with respect to the parameter λ. P 1 : (2λ + 5)x + (λ + 10)y + 3z 7 = 0, P 2 : 2x + (λ + 7)y + 2z + 15 = 0. 5.6 Projecting Planes A straight line lies in infinitely many planes. Any two of them determine the line. By eliminating one of the variables from two such equations we obtain an equation of a plane which contains the line and perpendicular to a coordinate plane. Thus if a line is given by the equations a 1 x + b 1 y + c 1 z + d 1 = 0 and a 2 x + b 2 y + c 2 z + d 2 = 0 then the elimination of y (for example) gives the plane Ax + Cz + D = 0 which is clearly perpendicular to xz-plane. The plane Ax + Cz + D = 0 is called the projecting plane for the given line. The equations Ax + Cz + D = 0, y = 0 determines a straight line in the projecting plane an xz-plane. This straight line is the projection of the given straight line on the xz-plane. a 1 x + b 1 y + c 1 z + d 1 = 0, a 2 x + b 2 y + c 2 z + d 2 = 0 xz-plane z intersection line projection of the intersection line O y x Figure 5.5: Projection of a line onto projecting plane 5.7 Intersection of three Planes We have the following possibilities: 1. All three planes are parallel. This includes the cases all planes coincide or two of them coincide. Generally parallelism of three planes means they have no point in common. 2. Two planes are parallel but the third one intersects these planes. If two planes coincide, there is a line of intersection of three planes. 3. No two planes are parallel but the pairwise intersection lines are parallel. This includes the case where the planes intersect in a single point. 4. The planes intersect in a single point.

5.7. INTERSECTION OF THREE PLANES 33 n 1 n2 n3 n1 n2 n3 n n2 3 n1 Three planes coincide. Two planes coincide and the third on is parallel to others. Three planes are parallel. Figure 5.6: First case n 1 n 3 n2 n 2 n 3 n 1 Figure 5.7: Second case Question 65. Discuss the intersection of given planes a) P 1 : 7x + 4y + 7z + 1 = 0, P 2 : 2x y z + 2 = 0, P 3 : x + 2y + 3z 1 = 0. b) P 1 : 2x y + 3z 5 = 0, P 2 : 3x + y + 2z 1 = 0, P 3 : 4x + 3y + z + 2 = 0. Question 66. Discuss the intersection of given planes with respect to the parameters a and b. P 1 : x + 2y z + b = 0, P 2 : 2x y + 3z 1 = 0, P 3 : x + ay 6z + 10 = 0.

34 CHAPTER 5. THE PLANE n 3 n 2 n 3 n 1 n 2 n1 Figure 5.8: Third case n 1 n2 n 3 Figure 5.9: Fourth case Question 67. Determine α such that the following four planes P 1 : x + 2y 3 = 0, P 2 : x + y 2z 9 = 0, P 3 : 3y + 5z + 15 = 0 and P 4 : 3x + αz 15 = 0 pass through the same point. Find the coordinates of this point. 5.8 Specialized Distance Formula In this section, we obtain rather specialized formulas for the distance from a point to a line and for distance between two skew lines. In both cases the distances are undirected. Let P 1 be a point on the line l and let m be the direction of l. Let P 0 be a point in R 3 not containing by the line l. Let α be the acute angle between m and P 1 P 0. It is clear that the distance d from P 0 to the line l is computed by the equation d = P 1 P 0 sin α. Using the properties of vector product we have P 1 P 0 m = P 1 P 0 m sin α and hence The equation 5.9 is called the specialized distance formula from a point to a line. d = P 1 P 0 m. (5.9) m

5.8. SPECIALIZED DISTANCE FORMULA 35 P 0 P 1 P 0 d P 1 α m l Figure 5.10: Specialized distance formula from a point to a line. The distance between parallel lines is similar to the above case. Now we want to obtain the smallest distance between two skew lines. Let P 1 and P 2 be any two points on the skew lines l 1 and l 2, respectively. Let m 1 and m 2 be directions of l 1 and l 2, respectively. It is clear that the projection of P 1 P 2 on m 1 m 2 is the line segment which has the length equal to the smallest distance between l 1 and l 2. The projection vector k is computed as P 1 P 2 ( m 1 m 2 ) k = m 1 m 2 2 ( m 1 m 2 ). The norm of the projection vector k is the perpendicular distance between l 1 and l 2. Hence the distance d is obtained by P 1 P 2 ( m 1 m 2 ) d = m 1 (5.10) m 2 l 1 l 2 P 1 P 2 m 2 m 1 m 1 m 2 θ P d Figure 5.11: Specialized distance between two skew lines. Question 68. Find the distance from (1, 0, 1) to the line l = {(x, y, z) : (x, y, z) = (2t 1, 2t + 2, t + 1), t R}. Question 69. Find the distance between the lines l 1 = {(x, y, z) : x + y z + 3 = 0, l 2 = {(x, y, z) : 3x + y + 5z 1 = 0, y + z + 2 = 0}. 2x y 4z + 2 = 0} and

36 CHAPTER 5. THE PLANE

Chapter 6 Transformation of Axes 6.1 Translation and Rotation of Axes If two sets of parallel coordinate axes are drawn, the origin O of the x y -coordinate plane has the coordinates α, β in the xy-coordinate system. Then a point P has the coordinates (x, y) in the xy-coordinate plane and (x, y ) in the x y -coordinate plane. It may be seen from the figure that two point of coordinates, that is, (x, y) and (x, y ) are related by the equations x = x + α, y = y + β. (6.1) y y P (x, y) (x, y ) α y x O β O x x Figure 6.1: Translation of axes. The equations 6.1 are called the equations of translation. One can obtain the coordinates of a point P in xycoordinate system using the coordinates of P in x y -coordinate system and the position of the origin O. Reverse is also true. Let us rotate the xy-coordinate axes about the origin through an angle θ to obtain the x y - coordinate axes. This time let us take a point P in the x y -coordinate plane and define its coordinates in the xy-plane. It is clear that the distance taken on the x-axis is equal to the projection on the x-axis of the distance taken on the x -axis minus the projection on the x-axis of the distance taken on the y -axis. Similarly, y-coordinate can be obtained and they can be combined in the equations x = x cos θ y sin θ, y = x sin θ + y cos θ. (6.2) 37

38 CHAPTER 6. TRANSFORMATION OF AXES y y P (x, y) x y y sin θ θ θ x x cos θ x Figure 6.2: Rotation of axes. The equations 6.2 are the equations of the rotation that gives the coordinates of a point in the xy-system after rotation about the origin through an angle θ. Question 70. a) Translate axes to the new origin (2, 3) and reduce the equation of the curve x 2 4y 2 4x+24y 36 = 0. b) For the equation x 2 + 4xy + y 2 = 1 rotate the axes through θ = 45. Question 71. The x, y axes are translated to x, y axes with origin O (2, 3). Find x, y coordinates of the point (4, 6) and transform the equation 4x 2 12y 2 16x 72y 128 = 0. Question 72. Find x, y coordinates of the point (9, 5) under the rotation of axes through the obtuse angle θ for which tanθ = 3 4 and transform the equation 3x 5y = 2.

Chapter 7 The Circle 7.1 The Equation of a Circle A circle is defined as a set of points in a plane which are at the same distance, the radius, from a fixed point, the center. Let C(α, β) be the center and r > 0 be the radius. Now let P (x, y) be an arbitrary (representative) point on the circle. Then we have CP = r. Hence using the distance formula (x α) 2 + (y β) 2 = r or (x α) 2 + (y β) 2 = r 2. (7.1) P (x, y) β C(α, β) α Figure 7.1: A circle with radius r and center C(α, β). The equation 7.1 is called the standard form of the equation of a circle and exhibits directly the radius and the center (α, β). Expending the equation 7.1 we get x 2 + y 2 2αx 2yβ + α 2 + β 2 r 2 = 0 x 2 + y 2 + Dx + Ey + F = 0. (7.2) The equation 7.2 is called the general form of the equation of a circle. It is second degree equation in x and y with no xy term and with equal coefficients of x 2 and y 2 terms. We can solve α, β and r in terms of E, D and F and get α = D 2, β = E 2, r = 1 2 D 2 + E 2 4F. (7.3) Therefore from a given general form we obtain the standard form as (x + D 2 )2 + (y + E 2 )2 = D2 + E 2 4F. 4 The equation 7.2 represents a circle if and only if D 2 + E 2 4F > 0 and it represents a single point if and only if D 2 + E 2 4F = 0 and it represents no real graph if and only if D 2 + E 2 4F < 0. If the general form represents a single point, the graph is sometimes called a point circle. In either equation (x α) 2 + (y β) 2 = r 2 or x 2 + y 2 + Dx + Ey + F = 0 three essential constants appear, α, β, r or E, D, F, respectively. This means that in general a circle may be required to satisfy three conditions which may be of various types. 39

40 CHAPTER 7. THE CIRCLE 7.2 Intersections Involving Circles We have three possible cases for the intersection of a straight line and a circle; 1. they have no point in common, 2. they have distinct two point in common, 3. they have exactly one point in common. In the third case, the line is tangent to the circle. We illustrate these cases in Figure 7.2. Figure 7.2: Possible intersections of a circle and a straight line. These cases may be characterized algebraically. If the equations of the line and the circle have the general forms ax+by +c = 0 and x 2 +y 2 +Dx+Ey +F = 0, respectively, then for the case b 0 we can find y as y = ax + c b and substitute this value into the general form of circle gives that x 2 + ( ax + c b ) 2 + Dx + E( ax + c ) + F = 0. b The result is a quadratic equation in x. If the roots of this quadratic equation in x are not real (imaginary) then the line and the circle do not meet. If the roots are real and distinct then there are two distinct points of intersection. Finally, if the roots are real and equal then there is one point of intersection, where the tangency appears. Let us now consider the intersection of two different circles C 1 and C 2 with general forms C 1 : x 2 + y 2 + D 1 x + E 1 y + F 1 = 0 and C 2 : x 2 + y 2 + D 2 x + E 2 y + F 2 = 0. If one equation is subtracted from the other, we get a linear equation in x and y. This is the line equation passing through the intersection of the circles C 1 and C 2. The straight line (D 1 D 2 )x + (E 1 E 2 )y + (F 1 F 2 ) = 0, which is obtained by subtracting the two circle equations of C 1 and C 2, is called the radical axis of the given circles C 1 and C 2. In general we have three cases, they have no common points or they have two distinct common points or they have exactly one point in common which is the point of tangency. Theorem 6. Let P (a, b) be a point on the radical axis of the circles C 1 and C 2. Then the lengths of the tangents from P to the points of tangency of the circles are equal.

7.2. INTERSECTIONS INVOLVING CIRCLES 41 Figure 7.3: Possible intersections of two circles P (a, b) l 1 l 2 r 1 r 2 C(α 1, β 1 ) C(α 2, β 2 ) x 2 + y 2 + E 2 x + D 2 y + F 2 = 0 x 2 + y 2 + E 1 x + D 1 y + F 1 = 0, (D 1 D 2 )x + (E 1 E 2 )y + (F 1 F 2 ) = 0 Figure 7.4: Radical axis of two circles. Proof. The figure 7.4 shows one possible case, the proof, however, applies to all cases. l 2 1 = (a α 1 ) 2 + (b β 1 ) 2 r 2 1 = a 2 + b 2 2aα 1 2bβ 1 + α 2 1 + β 2 1 r 2 1 l 2 2 = (a α 2 ) 2 + (b β 2 ) 2 r 2 2 = a 2 + b 2 2aα 2 2bβ 2 + α 2 2 + β 2 2 r 2 2 l 2 1 l 2 2 = 2(α 1 α 2 )a + 2(β 1 β 2 )b + (α 2 1 + β 2 1 r 2 1) (α 2 2 + β 2 2 r 2 2) l 2 1 l 2 2 = (D 1 D 2 )a + (E 1 E 2 )b + (F 1 F 2 ) = 0 l 2 1 = l 2 2 l 1 = l 2. Question 73. Find any points of intersection of the given curves and an equation of radical axis, and draw a picture. a) x 2 + y 2 + 2x = 0, x 2 + y 2 2y = 0. b) x 2 + y 2 + 8x 6y 11 = 0, x 2 + y 2 + 3x y 4. c) 2x 2 + 2y 2 + 4x 2y 1 = 0, x 2 + y 2 + 3x y 4 = 0.

42 CHAPTER 7. THE CIRCLE Question 74. Show that each of circles x 2 + y 2 = 9, x 2 + y 2 12x + 27 = 0 and x 2 + y 2 6x 8y + 21 = 0 is tangent to other two. Do the common tangents meet in a point? If they do, find the point. Question 75. Discuss the intersection followings with respect to the parameter λ. a) l : λx y + 1 = 0, C : x 2 + y 2 10x + 24 = 0. b) C 1 : x 2 + y 2 3x + 8y 5 = 0, C 2 : x 2 + y 2 + 6x 2y + λ = 0. 7.3 Systems of Circles Let the equations C 1 : x 2 + y 2 + D 1 x + E 1 y + F 1 = 0 and C 2 : x 2 + y 2 + D 2 x + E 2 y + F 2 = 0 represent two circles. The equation x 2 + y 2 + D 1 x + E 1 y + F 1 + k(x 2 + y 2 + D 2 x + E 2 y + F 2 ) = 0 (7.4) represents for each value of k, except 1, a circle through the points of intersection of the circles C 1 and C 2. The equation 7.4 also represents the family of circles passing through the intersection of C 1 and C 2. When k = 1, the equation 7.4 reduces to the equation of the radical axis. 7.4 The Angle between two Circles The angle between two curves is the angle between their tangents at the point of intersection. If the angle between two circles is π then the circles ar called orthogonal. 2 θ Figure 7.5: Angle between two circles. Question 76. Show that the circles C 1 : x 2 + y 2 6x 2y + 2 = 0 and C 2 : x 2 + y 2 4x + 4y + 6 = 0 intersect each other under right angle.