Topical Revision Paper 10 Unit 10 Circles

Topical Revision Paper 10 Term 3 Unit 10: Points, Lines and Shapes Question 1: The equation of a circle C1 is 22 20 4 100 0x y x y     . Find (i) the radius and the coordinates of the centre of the circle, [2] (ii) the value(s) of k if the circle touches the line yk . [2] (iii) the equation of the another circle, C2 that has the same centre as C1, and passes through the point ( 11,2) . [2] Solution: (i) 22 22 22 20 4 100 0 ( 10) 100 ( 2) 4 100 0 ( 10) ( 2) 4 x y x y xy xy                 Centre : ( 10,2) and Radius = 2 units (ii) Since the centre of the circle C1 is ( 10,2) and the radius is 2 units, k = 0 or 4 (iii) Radius of circle C2 = 1 unit Equation of circle C2 : 22( 10) ( 2) 1xy    [Standard Form] Or Equation of circle C2 : 22 20 4 103 0x y x y     [General Form] Question 2:

The diagram shows two circles C1 and C2 centred at A and B respectively. C1 passes through O and P and touches C2 at R. Q is on C2 such that QB is parallel to the y-axis. The length of PQ is 12 units and PQ is a tangent to both circles. Given that the equation of C1 is 22 18 0x y y   , find (i) the centre and radius of C1 , [3] (ii) the equation of C2 , [4] (iii) the equation of the perpendicular bisector of AB. [4] Answer: Solution: (i)           22 2222 22 2 1 1 18 0 18 9 0 9 0 9 9 The centre of : 0, 9 . The radius of 9 units. x y y x y y xy C C                (ii) y O A B P Q x R C1 C2 r 9 9  r

    2 22 2 22 Let the radius of be units. By Pythagoras ' Theorem, 9 12 9 81 18 144 81 18 36 144 4 Cr rr r r r r r r                       2 2 22 2 22 2 2 The radius of 4 units. The centre of : 12, 14 . 12 14 4 The equation of is 12 14 4 . C C xy C x y          (iii)

Question 3: The equation of a circle, C, is given by 22 4 10 0x y x y m     where m is a constant. A line 24yx is a tangent to the circle. (i) Write down the coordinates of the centre, O, of the circle. [1] (ii) Find the radius of the circle and the value of m. [5] (iii) The circle, C, is reflected in the line 24yx . Find the equation of the reflected circle. [3] Solution: (i) (2,5) 0 12 9 14Midpoint of , 22 1 6, 11 2 14 9Gradient of 12 0 5 12 Gradient of the perpendicular bisector of 12 5 2 2 5 1 AB AB AB y            121 2 625 1 12 2 11 142 5 5 12 9 25 5 10 The equation of the perpendicular bisector of is 12 9 25 . 5 10 x yx yx AB yx             

(ii) Since 24yx is a tangent to circle, 22 4 10 0x y x y m     22 2 2 22 (2 4) 4 10(2 4) 0 5 40 56 0 Discriminant = 0 40 (4)(5)(56 ) 0 480 20 0 24 Radius = 2 5 24 = 5 units x x x x m x x m m mm m                  (iii) 1 2 1 2 22 22 2 Since ( 4, 4) is the midpoint between two centres of the circles: = 5 and = 4 22 25 = 5 and = 4 22 (6, 3) is the centre of reflected circle Hence equation: ( 6) +( 3) = 10 or x x y y xy xy x    2 16 6 + 63 = 10 y x y   Question 4: (i) The equation of a circle is given by 22 10 2 1 0x y x y     . Find the centre and the radius of the circle. [2] (ii) The tangent to the circle at a point A is parallel to the straight line 20xy . Find the equation of the diameter of the circle through A. [2] (iii) The circle is then reflected about the y-axis. State the equation of the new circle formed. [1] Solution:

(i) Centre is (5, -1) and the radius is 225 ( 1) 1 5r      units (ii) The gradient of the straight line 20xy   is m = 1 so the gradient of the diameter through A is 1 . The equation of the diameter passes through A is 5 1( 1) 4 4 yx xy yx       (iii) The coordinates of the centre of the circle after the reflection about y-axis : (-5, -1) Hence the equation of the new circle formed is 2 2 2( 5) ( 1) 5xy   