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1 Solved Examples Example 1: Find the equation of the circle circumscribing the triangle formed by the lines x + y = 6, 2x + y = 4, x + 2y = 5. Method 1. Consider the equation (x + y 6) (2x + y 4) + λ 1 (2x + y 4) (x + 2y 5) + λ 2 (x + 2y 5) (x + y 6) = 0 (1) This equation is satisfied by the points of intersection of any two of the given three lines, i.e. it is satisfied by the vertices of the triangle formed by the given lines. Result: Now if (i) represents a circle then (a) coefficient of x 2 = coefficient of y 2 and (b) coefficient of xy = 0 And 3 + 5λ 1 + 3λ 2 = 0 λ 1 = 6/5 Substituting these values in (i) and simplifying We get x 2 + y 2 17x 19y + 50 = 0 (Ans.) Which is the equation of the required circle.

2 Method 2. Solve the lines in pair to find the vertices of the triangle and then obtain theequation of the circle through these three points. Example 2: y 2 = 4. Find the locus of the point of intersection of perpendicular tangents to the circle x 2 + Method 1. y = mx + 2 (1+m 2 ) is tangent to x 2 + y 2 = 4 for all values of m. It passes through (h, k) if k = mh + 2 (1+m 2 ) (k mh) 2 = 4 (1 + m 2 ) m 2 (h 2 4) 2h km + (k 2 4) = 0 (i) The roots m1 and m2 of this quadratic equation are the slopes of PT and PT. If PT and PT are at right angle then m1m2 = 1. From (i) m 1 m 2 = (k 2-4)/(h 2-4)

3 (k 2-4)/(h 2-4) = 1 h 2 + k 2 = 8 Locus of P(h, k) is x 2 + y 2 = 8 Method 2: Equation of tangent at the point T(2 cos θ, 2 sin θ) is 2 cos θ x + 2 sin θ y = 4 i.e. x cos θ + y sin θ = 2 (i) Tangent at T would be perpendicular to the tangent at T If TOT = 90o i.e. AOT = 90 + θ Co-ordinates of T are (2 cos (90 + θ), 2 sin (90 + θ)) Equation of tangent at the point T is 2 sin θ x + 2 cos θ y = 4 or, x sin θ + y cos θ = 2 (ii) Think: How to get the locus of point P(h, k)? Caution:

4 Do not simply square and add (i) and (ii). Though we get the required result, but that it not the right approach. Well, lines (i) and (ii) both pass through the point P(h, k) h cos θ + k sin θ = 2 and h sin θ + k cos θ = 2 Now we want to get a relation between h and k and also eliminate θ. The best way to do this is to square and ad above two Equations and we get h 2 + k 2 = 8 Locus of P(h, k) is x 2 + y 2 = 8 Method 3: OP 2 = OT 2 + TP 2 ( OTP is a right angled triangle) OP 2 = OT 2 + OT 2 OP 2 = 2R 2 h 2 + k 2 = 2(4) = 8 Locus of P(h, k) is x 2 + y 2 = 8. Note: x 2 + y 2 = 8 is director circle of the circle x 2 + y 2 = 4 Example 3: Find the condition that the line 3x + 44y p = 0 is tangent to the circle x 2 + y 2 4x 6y + 9 = 0 Radius of the given circle is 2 and centre is (2, 3). So for line 3x + 4x p = 0 to be

5 tangent to the circle we have, ( p)/ ( ) = p = p = 10 p = 8 18 p = 10 p = 28 Hence the required condition is p = 8 or p = 28. (Ans.) Example 4 A circular plot of land in the form of a unit circle is to be divided into two equal parts by the arc of a circle whose centre is on the circumference of the circular plot. Show that the radius of the circular arc is 2 cos θ, where θ is given by sin 2θ 2θ cos 2θ = π/2. Let O be the centre of the given circular plot of radius 1 i.e. OA = OB = OC = 1 and A be any point on its circumference. Again BDC be the arc of the circle with centre A and dividing the given circle into two equal parts. Let r the radius of the new circle, then AB = AC = AD = r. Let AOB = θ Then OBA = OCA = θ, and AOB = (π 2θ). Now area ABDCA must be = 1/2 area of unit circle = (π (1) 2 )/2 = π/2 (i) r/sin(π -2θ ) =1/(sin π ) r = 2 cos θ. Required Area APBDCQA = Area of sector ABDCA + area of sector OCQAPB 2 area of OAB

6 π/2 = 1/2 r 2 (2θ) + 1/2 (1) 2 (2π 4θ) 2 1/2 (1) 2 sin (π 2θ) sin 2θ 2θ cos 2θ = π/2 (Ans.) Example 5: Find the equation of the circle which passes through the point (2, 0) and whose centre is the limit of the point of intersection of lines 3x + 5y = 1, (2 + c)x + 5c 2 y = 1. Solving, 3x + 5y = 1, (2 + c)x + 5c 2 y = 1 We get, when c 1 Pause: We will study limits in detail in module 5. Now, we want to find out the equation of the circle which passes through (2, 0) and has its centre at (2/5), 1/25). Equation of the circle is (x-2/5) 2 +(y+1/25) 2 =(2/5-2) 2 +(1/25) 2 This is the required equation of the circle. Example 6: Find the equation of circle having the lines x 2 + 2xy + 3x + 6y = 0 as its normals and having size just sufficient to contain the circle, x(x 4) + y(y 3) = 0

7 The combined equation of two normal of the circle is given by x 2 + 2xy + 3x + 6y = 0 (x + 3)(x + 2y) = 0 x = 3, x = 2y Recall: A normal to a circle always passes through the centre of the circle. Now solving these, we get the co-ordinates of the centre of the circle as ( 3, 3/2); because the two normal intersect at the centre of the circle The required circle just contains the circle x(x 4) + y(y 3) = 0 i.e. x 2 + y 2 4x 3y = 0 (i) Hence the required circle will touch the circle given by (1) internally. Let r be the radius of the required circle. Now the two circles given by (1) = ([2 2 +(3/2) 2 ] )=5/2 and centre = (2,3/2) Now the required circle will touch the circle (i) internally, if We have distance between the centre of the two circles = difference between their radii. ((-3-2) 2 +(3/2-3/2) 2 )=(r-5/2) r = 15/2 Hence the equation of the required circle is given by (x + 3) 2 + (y-3/2) 2 =(15/2) 2 x 2 + y 2 + 6x 3y 45 = 0 (Ans.)

8 Example 7: A tangent is drawn to each of the circles x 2 + y 2 = a 2, x 2 + y 2 = b 2. Show that if these two tangents are perpendicular to each other, the locus of their point of intersection is a circle concentric with the given circles. Method 1: Let P (x 1, y 1 ) be the point of intersection of the tangents PA and PB where A, B are points of contact with the circles respectively. As PA is perpendicular to PB, the corresponding radii OA and OB are also perpendicular. Let AOX = θ BOX = 90 + θ Using the parametric from of the circle we can take A (a cos θ, a sin θ) B ( b sin θ, b cos θ) The equation of PA is x (a cosθ) + y (a sin θ) = a 2 y cos θ x sin θ = b Since P(x 1, y 1 ) lies on these tangents x 1 cos θ + y 1 sinθ = a and y 1 cos θ x 1 sin θ = b

9 as θ is a variable quantity; we eliminate θ. Squaring and adding above equation (we get). x y 1 2 = a 2 + b 2 locus of p is x 2 + y 2 = a 2 + b 2, which is concentric with given circles. Method 2: OAPB is a rectangle OP 2 = OA 2 + AP 2 x y 1 2 = a 2 + b 2 Locus of P(x 1, y 1 ) is x 2 + y 2 = a 2 + b 2 Example 9: The circle x 2 + y 2 = 1 cuts x-axis at P and Q. Another circle with centre at Q and variable radius intercepts the first circle at R above x-axis and the line segment PQ at S. Find the maximum area of the triangle QSR. Method 1. Equation of circle centred at Q is (x + 1) 2 + y 2 = µ 2 Since point R (cos θ, sin θ) lies on this circle µ 2 = (cosθ + 1) 2 + (sin θ) 2 = cos θ

10 = 2 (2cos 2 θ/2) µ = 2 cos θ/2 A = Area of QSR = 1/2 base altitude = 1/2 µ sin θ = cos θ/2 sin θ (i) da/dθ = cos θ cos θ/2-1/2 sin θ/2 sin θ Pause: We will study maxima and minima in module 5. For max./min. da/dθ = 0 tan θ = 2 cot θ/2 2t/(1-t 2 )=2/t (where t=tan θ/2) t = tan θ/2=1/ 2 (d 2 A)/(d 2 θ ) = sin θ cos θ/2-1/2 cos θ sin θ/2-1/2 sin θ/2 cos θ 1/4 cos θ/2 sin θ (A"(θ ) (at tan θ/2=1/ 2) ) = ve From (i) Maximum area = cos θ/2 sin θ = 2 sin θ/2 (cos θ/2) 2 4/(3 3) sq. units. (Ans.) Method 2.

11 Equation of circle I is x 2 + y 2 = 1 It cuts x-axis where P (1, 0) and Q ( 1, 0) Let QR = µ then equation of the circle II. Centred at Q( 1, 0) and radius = µ is given by (x + 1) 2 + y 2 = µ 2 (1) Solving it with x-axis; we get S (µ 1, 0). Also solving the two circles, we get the co-ordinates of R as [(µ 2 /2)-1,µ/2 ((4-µ 2 ) )] The area of QRS = 1/2 QS RL = 1/2 µ µ/2 ((4-µ 2 ) ) = A (say) Now A is max./min. means A 2 is max./min. Let A 2 = Z. Then Z = µ 4 /16 (4 µ) 2 dz/dµ=1/4. 4µ 3 (6µ 5 )/16 (d 2 z)/dµ 2 = 3µ 3 30 (µ 4 /16) For max./min. of A i.e. max./min. of A 2 or Z. we get dz/dµ = 0 µ = (8/3) and then

12 (d 2 z)/dµ 2 = 3.8/3-30/10.64/9 = ve For µ = ((8/3)) Thus area is max when µ = ((8/3)) Also max. area of QRS 1/4.8/3 ((4/3))=(4/3 3) sq. units Example 10: Distances from the origin to the centres of three circles x 2 + y 2 2λx = c 2 (where c is constant and λ is variable) are in G.P. Prove that the lengths of tangents drawn from any point on the circle x 2 + y 2 = c 2 to the three circles are in G.P. The equation of the three circles is x 2 + y 2 2λx = c 2 (1) where λ = λ 1, λ 2, λ 3. Their centres are: (λ 1, 0), (λ 2, 0) and (λ 3, 0) Now distances of these points from the origin are λ 1, λ 2 and λ 3 λ 1 λ 3 = λ 2 2 (2) Now, let P(h, k) be any point on the circle x 2 + y 2 = c 2, then h 2 + k 2 = c 2 (3) If r 1, r 2, r 3 are the lengths of the tangents from P(h, k) on The three circles, we then obtain r 1 2 = h 2 + k 2 2λ 1 h c 2 = c 2 2λh c 2, using (3) r 1 2 = 2λh

13 Similarly r 2 2 = 2λ 2 h r 3 2 = 2λ 3 h = r 1 2 r 3 2 = 4λ 1 λ 2 h 2 = 4λ 2 2 h 2 = (r 2 2) 2 = r 2 r 3 r 2 2 r 1, r 2, r 3 are in G.P. (Proved). Example 11: Show that the equation x 2 + y 2 4x ky 5 = 0 represents (for variable k) a family of circles passing through two fixed points A and B. Find the equation of the circle belonging to this family and cutting circle x 2 + y 2 6x 5y = 0 at right angles. x 2 + y 2 4x ky 5 = 0 (x 2 + y 2 4x 5) + k ( y) = 0 This is the equation of family of circles passing though the intersection point of x 2 + y 2 4x 5 = 0 (a circle) and a straight line putting y = 0, in x 2 + y 2 4x 5 = 0 gives x 2 4x 5 = 0 x = 1, 5 Hence the given circle passes through two fixed points ( 1, 0) and (5, 0) For given family of circles x 2 + y 2 4x ky 5 = 0 (1) g = 2, f = k/2, c = 5 One member of family (1) and circle x 2 + y 2 6x 5y 0 (2) Intersect orthogonally For circle (2) g = 3, f = 5/2, c = 0

14 For two circle to be orthogonal 2(gg + ff ) = (c + c ) 2[( 2) ( 3) + ( k/2)(5/2)] = [6+5K/4] = K/2 = 5 5K/2 = 17 K = 34/5 Required equation of circle is x 2 + y 2 4x + 34/5 y 5 = 0 (Ans.) Example 12: Lines 5x + 12y 10 = 0 and 5x 12y 40 = 0 touch a circle C 1 of diameter 6. If the centre of C 1 lies in first quadrant, find the equation of circle C 2 which is concentric with C 1 and cuts intercept of length 8 on these lines. Recall: If a circle touches two lines L 1 and L 2 then the centre of the circle lies on the angle bisectors of the lines. Angle bisector of given lines are (5x+12y-10)/13=±(5x-12y-40)/13 Taking +ve sign: y = (-5)/4 Taking ve sign: x = 5 Given lines L 1 and L 2 intersect at (5,-5/4) Since the centre of C 1 lies in the first quadrant, it can lie on x = 5 only.

15 Let the centre of C 1 be (5, y 1 ) (5(5)+12y 1-10)/ ( ) = y 1 = ± 39 y 1 = 2 or y 1 = -54/12 (Neglect) Centre of C 1 is (5, 2) Since C 2 is concentric with C 1, its centre is also (5, 2) C 2 cuts intercept of length 8 on lies 5x + 12y 10 = 0 and 5x 12y 40 = 0 AB = CD = 8 Recall: Perpendicular from centre bisects the chord. AN = 4 C 2 N = 3 (given) Radius of C 2 = r (say) r 2 = (AN) 2 + (C 2 N) 2 = r = 5

16 Equation of C 2 is (x 5) 2 + (y 2) 2 = 25 Example 13: Find the equation of the circle which passes through the point (2a, 0) and whose radical axis with respect to the circle x 2 + y 2 = a 2 is the lines x = a/2. Recall: Radical axis of two circles is S 1 S 2 = 0. Let equation of circle is x 2 + y 2 + 2gx + 2fy + c = 0 (i) radical axis of circle (i) and circle x 2 + y 2 a 2 = 0 is given by x 2 + y 2 + 2gx + 2fy + c = x 2 + y 2 a 2 2gx + 2fy + c + a 2 = 0 (ii) It is given that radical axis is x a/2 = 0 (iii) Comparing (ii) and (iii) We get f = 0, (c+a 2 )/2g=-a/2 ag + a 2 + c = 0 (iv) Circle (i) passes through (2a, 0) 4ag + 4a 2 + c = 0 (v) From (iv) and (v) 3ag + 3a 2 = 0 g = a

17 c = (ag + a 2 ) from (iv) = ( a 2 + a 2 ) = 0 Equation of circle is Example 14: x 2 + y 2 2ax = 0 (Ans.) Show that x 2 + y 2 + 4y 1 = 0, x 2 + y 2 + 6x + y + 8 = 0 and x 2 + y 2 4x 4y 37 = 0 touch each other. S 1 x 2 + y 2 + 4y 1 = 0 S 2 x 2 + y 2 + 6x + y + 8 = 0 S 3 x 2 + y 2 4x 4y 37 = 0 C 1 (0, 2), C 2 ( 3, 1/2), C 3 (2, 2) r 1 = (4+1) = 5 r 2 = (9+1/4-8)= 5/2 r 3 = (4+4+37)= 45=3 5 C 2 C 3 = ((2+3) 2 +(2+1/2) 2 )= (25+25/4)=5/2 5 C 1 C 3 = ((2-0) 2 +(2+2) 2 )=2 5

18 r 1 + r 2 = 3/2 5 = C 1 C 2 r 3 r 2 = 5/4 5 = C 2 C 3 r 3 r 1 = 2 5 = C 1 C 3 Recall: Two circles touch each other. (i) externally if C 1 C 2 = r 1 + r 2 (ii) internally if C 1 C 2 = r 1 r 2 Example 15: Find the four common tangents to the circles x 2 + y 2 22x + 4y = 0 and x 2 + y 2 22x 4y 100 = 0 S 1 x 2 + y 2 22x + 4y = 0 S 2 x 2 + y x 4y 100 = 0 C 1 (11, 2), C 2 ( 11, 2)

19 r 1 = ((11) 2 +(-2) 2-100) = 5 r 2 = ((-11) 2 +(2) ) = 15 Out of four common tangents are transverse tangents and other two direct tangents. (1) and (2) are direct common tangents while (3) and (4) are transverse common tangents. Recall: Transverse common tangents divide line joining centres in ratio of radii internally while direct tangents divides line joining centres in ratio of radii externally. Let T 1, T 2 divide C 1 C 2 in ratio of r 1 : r 2 internally and externally respectively. Co-ordinates of T 1 are ( (-11))/(15+5) and (15 (-2)+5 2)/(15+5) That is T 1 is the point (11/2,-1) Co-ordinates of T 2 are ( (-11))/(15-5) and (15 (-2)-5 2)/(15-5) that is T 2 is the point (22, 4) Let the equation to either of the tangents, passing through T 1 be y + 1 = m (x 11/2) (A) Then the perpendicular from the point (11, 2) on it is equal to + 5 and hence (m(11-11/2)-(-2+1))/ (1+m 2 ) = ± 5 On solving, we have m = -24/7 or 4/3

20 The required tangents through T 1 are therefore 24x + 7y = 125, and 4x 3y = 25 Similarly the equation to the tangents through T 2 is y + 4 = m (x 22) (B) where (m(11-22)-(-2+4))/ (1+m 2 ) = ± 5 On solving, we have m = 7/24 or 3/4 On substitution in (B) the required equations are therefore x 24y = 250 and 3x + 4y = 50 The four common tangents are therefore found. (Ans.) Example 16: The circle x 2 + y 2 4x 4y + 4 = 0 is inscribed in a triangle which has two of its sides along the co-ordinate axes. The locus of the circumcentre of the triangle is x + y xy + k (x 2 + y 2 ) 1/2 = 0 Find k? Method 1. The equation of the incircle can be put in the from (x 2) 2 + (y 2) 2 = 4 This implies that the inradius r = 2 (i) Let the hypotenuse of the triangle meet OX and OY at A (a, 0) and B (0, b) respectively. r = (Area of AOB) / ((1/2)(Sum of sides of AOB)) = ((1/2)ab) / ((1/2)(a+b+ (a 2 +b 2 ))) 2 = ab/(a+b+ (a 2 +b 2 )) Let M (x 1, y 1 ) be the circumcentre of OAB. Since OAB is right angled, it s circumcentre

21 is the mid point of hypotenuse. 2 = (4x 1 y 1 ) / (2x 1 +2y 1 +2 (x 1 2 +y 1 2 )) x 1 + y 1 + (x 1 2 +y 1 2 ) = x 1 y 1 Hence the equation of the locus of M(x 1, y 1 ) is x + y xy + (x 2 +y 2 ) = x 1 y 1 Comparing it with the given equation of the locus, we find that k = 1. (Ans.) Method 2. Equation of AB is x/a + y/b = 1 (1) (x 1, y 1 ) (a/2,b/2), where M(x 1, y 1 ) is the circumcentre of OAB i.e. midpoint of the hypotenuse. (1) becomes: x / (2x 1 )+y / (2y 1 ) = 1 (2) In circle (x 2) 2 + (y 2) 2 = 4 touches line (2) (2/(2x 1 )+2 / (2y 1 )-1) / ((1/2x 1 ) 2 + (1/2y 1 ) 2 ) = 2 x 1 + y 1 x 1 y 1 + (x 1 2 +y 1 2 ) = 0 Locus of M(x 1, y 1 ) is x + y xy + (x 2 +y 2 ) = 0 Comparing it with the given equation of the locus, we find that k = 1. Note: Distance from (2, 2) to the line x/2x 1 +y/2y 1-1 = 0 has been taken 2, because origin and this point lies on the same side of the origin.

+ 2gx + 2fy + c = 0 if S

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