b UVW is a right-angled triangle, therefore VW is the diameter of the circle. Centre of circle = Midpoint of VW = (8 2) + ( 2 6) = 100

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1 Circles 6F a U(, 8), V(7, 7) and W(, ) UV = ( x x ) ( y y ) = (7 ) (7 8) = 8 VW = ( 7) ( 7) = 64 UW = ( ) ( 8) = 8 Use Pythagoras' theorem to show UV UW = VW 8 8 = 64 = VW Therefore, UVW is a right-angled triangle. b UVW is a right-angled triangle, therefore VW is the diameter of the circle. Centre of circle = Midpoint of VW 7 ( ) 7 ( ), = (, ) c Radius of the circle is of VW = = 4 = 4 (x ) (y ) = 4 a A (, 6), B(5, 7) and C(8, ) Use Pythagoras' theorem to show AB BC = AC AB = ( x x) ( y y) = (5 ) (7 6) = 0 BC = (8 5) ( 7) = 90 AC = (8 ) ( 6) = 00 Therefore, ABC is a right-angled triangle and AC is the diameter of the circle. b Centre of circle = Midpoint of AC 8 6 ( ), = (5, ) Radius of the circle is of AC = 00 (x 5) (y ) = 5 = 5 c Base of triangle = AB = 0 units Height of triangle = BC = 90 units Area of triangle ABC = 0 90 = 5 units Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

2 a i A(, 9) and B(9, ) 9 9, = (, 5) y y The gradient of the line segment AB = x x 9 = = 9 ( ) So the gradient of the line perpendicular to AB is. y y= mx ( x) m = and ( x, y ) = (, 5) So y 5 = (x ) y = x ii A(, 9) and C(5, ) 5 9, = (9, 0) y y 9 The gradient of the line segment AC = = x x 5 So the gradient of the line perpendicular to AC is. y y = mx ( x) m = and ( x, y ) = (9, 0) So y 0 = (x 9) y = x 4 = b Solve y = x and y = x 4 simultaneously x = x 4 9x 6 = 4x 4 x = 9 x =, y = () 4 = 6 So, the coordinates of the centre of the circle are (, 6) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

3 c Radius = distance from (, 6) to (9, ) ( x x ) ( y y ) = (9 ) ( 6) = 5 = (x ) (y 6) = 69 4 a i P(, 8) and Q(6, 7) 6 8 7, = 7, y y The gradient of the line segment PQ = x x = 78 = 6 So the gradient of the line perpendicular to PQ is. y y = mx ( x) m = and 7 ( x, y ) =, So y = 7 x y = x 0 ii Q(6, 7) and R(4, 7) QR is the line y = , = (, 7) x =. b Solve y = x 0 and x = simultaneously to find the centre of the circle: () 0 = y y = The centre of the circle is (, ) Radius = distance from (, ) to (4, 7) ( x x) ( y y) = (4 ) ( 7 ) = 5 (x ) (y ) = 5 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free.

4 5 R(, ) and S(4, ) 4, = (, ) y y The gradient of the line segment RS = = x x 4 = So the gradient of the line perpendicular to RS is. y y= mx ( x) m = and ( x, y ) = (, ) So y = (x ) y = x 5 S(4, ) and T(0, 5) 4 0 5, = (7, ) y y The gradient of the line segment ST = x x = = 4 So the gradient of the line perpendicular to ST is 4. y y= mx ( x) m = 4 and ( x, y ) = (7, ) So y = (x 7) 4 y = 4 x 5 4 Solve y = x 5 and y = 4 x 5 4 simultaneously x 5 = 4 x 5 4 x 0 = x 5 5x = 45 x =, y = () 5 = 4 So the centre of the circle is (, 4) Radius = distance from centre (, 4) to (, ) ( x x ) ( y y ) = ( ) ( 4) = 50 The equation of the circle is (x ) (y 4) = 50 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 4

5 6 a A(, 5), B(, ) and C(7, 5) Using Pythagoras' theorem AB BC = AC AB = ( x x ) ( y y ) = ( ) ( 5) = = 400 BC = ( x x ) ( y y ) = (7 ) (5 ) = 6 64 = 00 AC = ( x x) ( y y) = (7 ) (5 5) = = 500 Therefore ABC is a right-angled triangle b Centre of circle = midpoint of AC =, = (, 5) Radius = of AC = 0 5 of 500 = = 5 5 Equation of circle: (x ) (y 5) = (5 5) or (x ) (y 5) = 5 c We know that A, B and C all lie on the circumference of the circle. D(8, 0), substitute x = 8 and y = 0 into the equation of the circle: (8 ) (0 5) = 00 5 = 5 Therefore, D(8, 0) lies on the circumference of the circle (x ) (y 5) = 5 7 a A(, 9), B(6, 0), C(7, ), D(0, ) The length of AB is ( x x ) ( y y ) ( ( )) ( ) = = 7 = 49 = 50 The length of BC is ( x x ) ( y y ) ( ) ( ) ( ) = = 7 = 49 = 50 The length of CD is ( x x ) ( y y ) ( ) ( ) ( ) ( ) = 0 7 = 7 = 49 = 50 The length of DA is ( x x ) ( y y ) ( ) ( ) ( ) = 0 9 = 7 = 49 = 50 The sides of the quadrilateral are equal. The gradient of AB is y y 0 9 = = x x 6 7 ( ) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 5

6 7 a The gradient of BC is y y 0 7 = = = 7 x x 76 The product of the gradients = 7 =. 7 So the line AB is perpendicular to BC. So the quadrilateral ABCD is a square. b The area = = 50 c The mid-point of AC is x x y y 7 9 6, =, =, =, 6 So the centre of the circle is (, 6). 8 a D(, ), E(0, b), F(, 5) Using Pythagoras' theorem DE EF = DF DE = ( x x) ( y y) = (0 ) (b ) = 4 b 6b 9 = b 6b EF = ( x x) ( y y) = ( 0) (5 b) = 44 b 0b 5 = b 0b 69 DF = ( x x) ( y y) = ( ) (5 ) = 96 4 = 00 b 6b b 0b 69 = 00 b 6b 8 = 0 b 8b 9 = 0 (b 9)(b ) = 0 b = 9 or b = As b > 0, b =. ( ) 5 b Centre of circle = midpoint of DF =, = (5, 4) Distance of radius = of DF = 0 of 00 = = 5 Equation of circle: (x 5) (y 4) = (5 ) = 50 Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 9 a x x y 4y 4 = 0 Completing the square gives: (x ) (y ) 44 4 = 0 (x ) (y ) = 69 Centre of the circle is (, ) and the radius of the circle is. b If AB is the diameter of the circle then the midpoint of AB is the centre of the circle. 7 7 Midpoint of AB =, = (, ) Therefore, AB is the diameter of the circle. c The point C lies on the x-axis, so y = 0. Substitute y = 0 into the equation of the circle. (x ) (0 ) = 69 x x 44 = 69 x x 4 = 0 (x 6)(x 4) = 0 x = 6, x = 4 As x is negative, x = 6 The coordinates of C are (6, 0) Pearson Education Ltd 07. Copying permitted for purchasing institution only. This material is not copyright free. 7

Circles, Mixed Exercise 6

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