急! F5 Equations of circles 8q1

14.
(a)
Let x² + y² + Dx + Ey + F = 0 be the equation of the circle.

O(0, 0), P(2, 4) and Q(8, -4) lie on the circle :
(0)² + (0)² + D(0) + E(0) + F = 0 ...... [1]
(2)² + (4)² + D(2) + E(4) + F = 0 ...... [2]
(8)² + (-4)² + D(8) + E(-4) + F = 0 ...... [3]

From [1] :
F = 0

Put F = 0 into [2] :
4 + 16 + 2D + 4E = 0
2D + 4E = -20 ...... [4]

Put F= 0 into [3] :
64 + 16 + 8D - 4E = 0
2D - E = -20 ...... [5]

[4] - [5] :
5E = 0
E = 0

Put E = 0 into [5] :
2D = -20
D = -10

Hence, the equation of the circle : x² + y² - 10x = 0

(b)
When x = 5 and y = 5 :
L.S.
= (5)² + (5)² - 10(5)
= 0
= R.S.

R(5, 5) lies on the circle.
Since all of O, P, R and Q lie on the circle, then PORQ is a cyclic quadrilateral.


15.
Let C(a, b) be the centre of the circle.
a < 0 and b < 0 for C is in the 3rd quadrant.

CE = CF
√[(a - 0)² + (b - 3)²] = √[(a - 0)² + (b + 5)²]
a² + (b - 3)² = a² + (b + 5)²
b² - 6b + 9 = b² + 10b + 25
16b = -16
b = -1

CE = √41
√[(a - 0)² + (b - 3)²] = √41
a² + (b - 3)² = 41 ...... [1]

Put b = -1 into [1] :
a² + (-1 - 3)² = 41
a² - 25 = 0
(a + 5)(a - 5) = 0
a = -5 or a = 5(rejected)

Centre C = (-5, -1)

Equation of the circle :
(x + 5)² + (y + 1)² = (√41)²
x² + 10x + 25 + y² + 2y + 1 = 41
x² + y² + 10x + 2y - 15 = 0