Circles, Mixed Exercise 6 a QR is the diameter of the circle so the centre, C, is the midpoint of QR ( 5) 0 Midpoint = +, + = (, 6) C(, 6) b Radius = of diameter = of QR = of ( x x ) + ( y y ) = of ( 5 ) + (0 ) = = of 00 of 0 = 0 units c Circle with centre (, 6) and radius 0: (x ) + (y 6) = 00 d P(, 6) lies on the circle if P satisfies the equation, so substitute x = and y = 6 into the equation of the circle: ( ) + (6 6) = 00 + 0 = 00 Therefore, P lies on the circle. The distance between (0, 0) and (5, ) is ( x x ) ( y y ) ( ) ( ) ( ) + = 5 0 + 0 = 5 + = 5 + = 9 The radius of the circle is 0. As 9 < 0 (0, 0) lies inside the circle. a x + x + y + 6y = x y 7 x + y + 8y = 7 Completing the square gives: (x 0) + (y + ) 6 = 7 (x 0) + (y + ) = 9 Centre of the circle is (0, ) and the radius is. b The circle intersects the y-axis at x = 0 (0 0) + (y + ) = 9 y + 8y + 6 = 9 y + 8y + 7 = 0 (y + )(y + 7) = 0 y = or y = 7 (0, ) and (0, 7) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
c At the x-axis, y = 0 x + 0 + 8(0) = 7 x = 7 There are no real solutions, so the circle does not intersect the x-axis. a The centre of ( ) ( ) x 8 + y 8 = 7 is (8,8). Substitute (8, 8) into ( x+ ) + ( y ) = ( 8 ) ( 8 ) 9 5 8 5 06 So (8, 8) lies on the circle ( x ) ( y ) 06 + + = + = + = + + = 06. b As Q is the centre of the circle ( x ) ( y ) length PQ must equal the radius. So PQ = 06 + + = 06 and P lies on this circle, the Alternative method: Work out the distance between P(8, 8) and Q(, ) using the distance formula. 5 a Substitute (, 0) into ( ) + ( 0) = + 0= So (, 0) is on the circle. x + y = Substitute, into x + y = + = + = So, is on the circle. Substitute, into x + y = + = + = So, is on the circle. Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
5 b The distance between (, 0) and, is + = + 0 ( x x ) ( y y ) ( ) = + + = + 9 = + = = The distance between (, 0) and, is + = + 0 ( x x ) ( y y ) ( ) = + + = + 9 = + = = Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
5 b The distance between, and, is + = + ( x x ) ( y y ) = + ( ) 0 = 0+ So AB, BC and AC all equal. ABC is equilateral. = 6 a (x k) + (y k) =, (, 0) Substitute x = and y = 0 into the equation of the circle. ( k) + (0 k) = 9 6k + k + 9k = 0 0k 6k = 0 5k k = 0 (5k + )(k ) = 0 k = or k = 5 b As k > 0, k = Equation of the circle is (x ) + (y ) = 7 x + px + y + y = 0, y = x 9 Substitute y = x 9 into the equation x + px + y + y = 0 x + px + (x 9) + (x 9) = 0 x + px + 9x 5x + 8 + x 6 0 = 0 0x + (p )x + 5 = 0 There are no solutions, so using the discriminant b ac < 0: ( p ) ( 0)( 5) < 0 ( p ) < 000 p <± 000 p < ± 000 p < ± 0 0 0 0 < p < + 0 0 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
8 Substitute x = 0 into y = x 8 y = (0) 8 y = 8 Substitute y = 0 into y = x 8 0 = x 8 x = 8 x = The line meets the coordinate axes at ( 0, 8) and ( ),0. The coordinates of the centre of the circle are at the midpoint: x+ x y+ y 0+ 8+ 0 8, =, =, =, The length of the diameter is ( ) ( x x ) ( y y ) ( ) ( ) ( ) + = 0 + 0 8 = + 8 = 80 = 6 5 = 5 So the length of the radius is 5 5. = The centre of the circle is (, ) The equation of the circle is x x + y y = r ( ) ( ) ( x ) y ( ) ( x ) ( y ) ( ) ( ) + = 5 9 a The radius is b + + = 0 and the radius is 5. ( x x ) ( y y ) ( ) ( ) + = 8 + 0 0 = + 0 = 6 + 00 = 6 = 9 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5
9 b The centre is on the perpendicular bisector of (,0 ) and (,0.) + a = 8 + a = 6 a = 0 Substitute y = 0 into ( x ) y ( x 5) = 6 x 5 = 6 x 5=± 6 So x 5 = 6 x= and x 5 =6 x= 5 + = 6, 0. The coordinates of P and Q are (, 0) and ( ) Substitute x = 0 into ( x ) ( y ) + ( y 7) = 6 + ( y 7) = ( y 7) = 05 y 7 =± 05 So y = 7 ± 05 + + 7 = The values of m and n are 7 + 05 and 7 05. a (x + 5) + (y + ) = 5, A(a, 0), B(0, b) At A(a, 0): (a + 5) + (0 + ) = 5 a + 0a + 5 + 5 = 0 a + 0a 96 = 0 (a + 6)(a 6) = 0 As a > 0, a = 6 At B(0, b): (0 + 5) + (b + ) = 5 5 + b + b + 5 = 0 b + b 96 = 0 (b + )(b 8) = 0 As b > 0, b = 8 So a = 6, b = 8 a So Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
b A(6, 0), B(0, 8) y y gradient = x x y-intercept = 8 = 8 0 = 0 6 Equation of the line AB is y = x + 8 c Area of triangle OAB = 6 8 = units a By symmetry p = 0. Using Pythagoras theorem q q + 7 = 5 + 9 = 65 q = 576 q = ± 576 q = ± As q > 0, q =. b The circle meets the y-axis at q± r; i.e. at + 5 = 9 and 5 = So the coordinates are (0, 9) and (0, ). Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
The gradient of the line joining (, 7) and ( ) y y ( 7) + 7 8 = = = = x x 5( ) 5+ 8 5 So the gradient of the tangent is The equation of the tangent is y y = m xx ( ) ( 7) x ( ) y+ 7= ( x+ ) ( ) y = y+ 7=x y =x 0 or x+ y+ 0 = 0 =. ( ) 5, is Let the coordinates of C be (p, q). (, ) is the mid-point of (, 7) and (p, q) So + p and 7 + q = = + p = + p = p = 7 + q = 7+ q = q = 9 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8
5 So the coordinates of C are (, 9). The length of AB is ( x x ) ( y y ) ( ) ( ) ( ) ( ) + = 5 + 7 = 8 + = 6 + 6 = 80 The length of BC is ( x x ) ( y y ) ( ) ( ) ( ) ( ) ( ) + = 5 + 9 = 6 + = 6 + = 80 The area of ABC is 80 80 = 00 = 00 = 00 = 0 = 60 6 (x 6) + (y 5) = 7 Centre of the circle is (6, 5). Equation of the line touching the circle is y = mx + Substitute the equation of the line into the equation of the circle: (x 6) + (mx + 7) = 7 x x + 6 + m x + mx + 9 7 = 0 ( + m )x + (m )x + 68 = 0 There is one solution so using the discriminant b ac = 0: (m ) ( + m )(68) = 0 96m 6m + 7m 7 = 0 76m + 6m + 8 = 0 9m + 8m + = 0 (9m + 8)(m + ) = 0 8 m = or m = 9 8 y = x + and y = x + 9 y y 7 a Gradient of AB = = 7 = x x 5 + 5 7+ Midpoint of AB =, = (, ) M(, ) Line l is perpendicular to AB, so gradient of line l = y y= mx ( x) y = (x ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 9
7 a y = x + 8 b C(, c) y = () + 8 = C(, ) Radius of the circle = distance CA = ( x x) + ( y y) = Equation of the circle is (x + ) + (y ) = 50 ( ) + ( 7) = 50 c Base of triangle = distance AB = Height of triangle = distance CM = (5 ) + ( 7) = 0 ( + ) + ( ) = 0 Area of triangle ABC = 0 0 = 0 units 8 a (x ) + (y + ) = 5 The equations of the lines l and l are y = x + c Diameter of the circle that touches l and l has gradient and passes through the centre of the circle (, ) y = x + d = () + d d = y = x is the equation of the diameter that touches l and l. Solve the equation of the diameter and circle simultaneously: (x ) + ( x + ) = 5 x 6x + 9 + 9 x 8 x + 5 = 0 9 x 6 x 9 = 0 x 78x 5 = 0 x 6x 7 = 0 (x 9)(x + ) = 0 x = 9 or x = Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 0
8 a When x = 9, y = (9) = 7 When x =, y = () = (9, 7) and (, ) are the coordinates where the diameter touches lines l and l. P(9, 7) and Q(, ) b The equations of the lines l and l are y = x + c l touches the circle at (, ): = () + c, c =, so y = x + l touches the circle at (9, 7): 7 = (9) + c, c =, so y = x ` 9 a x + 6x + y y = 7 Equation of the lines are y = mx + 6 Substitute y = mx + 6 into the equation of the circle: x + 6x + (mx + 6) (mx + 6) = 7 x + 6x + m x + mx + 6 mx 7 = 0 ( + m )x + (6 + 0m)x + 7 = 0 There is one solution so using the discriminant b ac = 0: (6 + 0m) ( + m )(7) = 0 00m + 0m + 6 68m 68 = 0 m + 0m = 0 m + 5m = 0 (m )(m + ) = 0 m = or m = y = x + 6 and y = x + 6 b The gradient of l = and the gradient of l =, so the two lines are perpendicular, Therefore, APRQ is a square. x + 6x + y y = 7 Completing the square: (x + ) 9 + (y ) = 7 (x + ) + (y ) = 7 Radius = 7 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
9 b Let point P have the coordinates (x, y) Using Pythagoras' theorem: l: (0 x) + (6 y) = 7 Using the equation for l, y = x + 6, ( 0 x) + 6 x+ 6 = 7 x + x = 7 6 7 x = 7 6 x = 6 x = ± From the diagram we know that x is negative, so x =, y = () + 6 = 5 P(, 5) Now let point Q have the coordinates (x, y). Using the equation for l, y = x + 6, (0 x) + (6 (x + 6)) = 7 x + 6x = 7 7 x = 7 x = x = ± From the diagram x is positive, so x =, y = () + 6 = Q(, ) c Area of the square = radius = 7 units 0 a Equation of the circle: (x 6) + (y 9) = 50 Equation of l: y = x + Substitute the equation of the line into the equation of the circle: (x 6) + (x + ) = 50 x x + 6 + x x + 50 = 0 x 6x + 0 = 0 x 8x + 65 = 0 (x )(x 5) = 0 x = or x = 5 When x =, y = + = 8 When x = 5, y = 5 + = 6 P(5, 6) and Q(, 8) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
y y 0 b The gradient of the line AP = x x = 9 6 6 5 = 7 So the gradient of the line perpendicular to AP, l, is 7. The equation of the perpendicular line is y y= mx ( x) m = 7 and (, ) x y = P(5, 6) So y 6 = (x 5) 7 y = 7 x + 07 7 y y The gradient of the line AQ = = 9 8 = x x 6 7 So the gradient of the line perpendicular to AQ, l, is 7. The equation of the perpendicular line is y y = mx ( x) m = 7 and ( x, y ) = Q(, 8) So y 8 = 7(x ) y = 7x 8 l: y = 7 x + 07 and l: y = 7x 8 7 y y c The gradient of the line PQ = = 8 6 = x x 5 So the gradient of the line perpendicular to PQ, l, is. The equation of the perpendicular line is y y = mx ( x) m = and ( x, y ) = A(6, 9) So y 9 = (x 6) l: y = x + d l: y = 7 x + 07, l: y = 7x 8 and l: y = x + 7 Solve these equations simultaneously one pair at a time: l and l: 7 x + 07 = 7x 8 7 x + 07 = 9x 58 8x = 688 x =, so y = 7 8 = 5 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
5 0 d l and l intersect at,. l and l: 7x 8 = x + 6x = 86 x =, so y = + = 5 Therefore all three lines intersect at R 5, e Area of kite APRQ = AR PQ PQ = AR = ( x x ) + ( y y ) = ( x x ) ( y y ) Area = 5 a y = x + ( 5) + (8 6) = 8 = 8 5 + = ( 6 ) ( 9 ) 8 = 00 Substitute x = 0 into y = x + y = ( 0) + = So A is (0, ) Substitute y = 0 into y = x + 0 = x + x = x = So B is (, 0). b The mid-point of AB is x+ x y+ y 0 + + 0, =, =,6 c + = ( ) 5 5 + = 50 9 = 5 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.
c AOB = 90, so AB is a diameter of the circle. The centre of the circle is the mid-point of AB, i.e. (, 6). The length of the diameter AB is ( x x ) ( y y ) ( ) ( ) ( ) + = 0 + 0 = + = 6 + = 60 So the radius of the circle is The equation of the circle is 60 + = 60 + = ( x ) ( y 6) ( x ) ( y 6) ( x ) ( y ) + 6 = 0 60. a A(, ), B(6, 0) and C(, q) AB = ( x x ) + ( y y ) = ( 6 + ) + (0 + ) = Diameter = BC = ( 6) ( 0) + + q = 9 + q AC = + + q + = ( ) ( ) + + + = 6 q q q + q+ 0 Using Pythagoras' theorem AC + AB = BC : (q + q + 0) + = 9 + q q 6 = 0 q = b The centre of the circle is the midpoint of B(6, 0) and C(, ) 6 0 Midpoint BC = +, + = 5, The radius is half of BC = of 9 + q = of 9 + = of 65 = 65 5 65 Equation of the circle is x+ + ( y ) = 5 65 x+ + ( y ) = Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5
a R(, ), S(7, ) and T(8, 7) RT = (8 + ) + ( 7 ) = RS = (7 + ) + ( ) = ST = (8 7) + ( 7 ) = Using Pythagoras' theorem, ST + RS = + = = RT, therefore, RT is the diameter of the circle. b The radius of the circle is diameter = = 6 = 6 = 6 = 6 The centre of the circle is the mid-point of RT: x+ x y+ y + 8 + ( 7), =, =, =, So the equation of the circle is ( x ) ( y ) ( ) ( x ) ( y ) + + = 6 or + + = 6 ( ) A(, 0), B(, 8) and C(6, 0) y y The gradient of the line AB = = 8 0 = x x + So the gradient of the line perpendicular to AB, is. 0 8 Midpoint of AB = +, + = (0, ) The equation of the perpendicular line through the midpoint of AB is y y = mx ( x) m = and ( x, y ) = (0, ) So y = (x 0) y = x + y y The gradient of the line BC = x x = 0 8 6 = So the gradient of the line perpendicular to BC, is. Midpoint of BC = + 6 8+ 0, = (5, ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6
The equation of the perpendicular line through the midpoint of BC is y y= mx ( x) m = and (, ) x y = (5, ) So y = (x 5) y = x + Solving these two equations simultaneously will give the centre of the circle: x+ = x+ x+ 6 = x+ 5x = 5 x =, so y = + = The centre of the circle is (, ). The radius is the distance from the centre of the circle (, ) to a point on the circumference C(6, 0): Radius = (6 ) + (0 ) = The equation of the circle is (x ) + (y ) = 5 a i A(7, 7) and B(, 9) y y The gradient of the line AB = = 9 7 = x x + 7 So the gradient of the line perpendicular to AB, is. 7 7 9 Midpoint of AB = +, + = (, 8) The equation of the perpendicular line is y y = mx ( x) m = and ( x, y ) = (, 8) So y 8 = (x + ) y = x ii C(, ) and D(7, ) The line CD is y = 7 + Midpoint of CD =, = (, ) The equation of the perpendicular line is x = Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7
5 b The two perpendicular bisectors cross at the centre of the circle Solve y = x and x = simultaneously: y = () = The centre of the circle = (, ) The radius is the distance from the centre of the circle (, ) to a point on the circumference C(, ): Radius = ( + ) + ( ) = The equation of the circle is (x + ) + (y ) = Challenge a Solve (x 5) + (y ) = 0 and (x 0) + (y 8) = 0 simultaneously: x 0x + 5 + y 6y + 9 = 0 and x 0x + 00 + y 6y + 6 = 0 x 0x + y 6y + = 0 and x 0x + y 6y + 5 = 0 x 0x + y 6y + = x 0x + y 6y + 5 0x 6y + = 0x 6y + 5 0x + 0y = 0 x + y = x + y = 0 b Solve (x 5) + (y ) = 0 and x + y = simultaneously: (9 y) + (y ) = 0 8 8y + y + y 6y + 9 = 0 y y + 70 = 0 y y + 5 = 0 (y 5)(y 7) = 0 y = 5 or y = 7 When y = 5, x = 5 = 9 When y = 7, x = 7 = 7 P(7, 7) and Q(9, 5) c Area of kite APBQ = PQ AB PQ = (9 7) + (5 7) = 8 A(5, ) and B(0, 8) AB = (0 5) + (8 ) = 50 Area = 8 50 = 00 = 0 units Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 8