Form Five

A circle passes through the origin and cuts the axes a t (3,0) and (0, -2) respectively. Find the equation if the circle.

Let the circle be: x2 + y2 + Dx + Ey + F = 0

(0,0) lies on the circle:
(0)2 + (0)2 + D(0) + E(0) + F = 0
Hence, F = 0

(3,0) lies on the circle:
(3)2 + (0)2 + D(3) + E(0) + F = 0
Hence, D = -3

(0,-2) lies on the circle:
(0)2 + (-2)2 + D(0) + E(-2) + F = 0
Hence, E = 2

The equation of the circle is:
x2 + y2 - 3x + 2y = 0

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Find the equation of the circle with centre on the line 2x + y =0 and passing through the points A (2, - 1) and B (0, - 3)

Let O(a, b) be the centre of the circle.

OA = OB
√[(a - 2)2 + (b + 1)2] = √[(a - 0)2 + (b + 3)2]
(a - 2)2 + (b + 1)2 = (a - 0)2 + (b + 3)2
a2 - 4a + 4 + b2 + 2b + 1 = a2 + b2 + 6b + 9
-4a - 4b = 4
a + b = -1 ...... (1)

O(a, b) lies on 2x + y = 0
2a + b = 0 ...... (2)

(1)-(2):
-a = -1
a = 1

Put a = 1 into (1):
(1) + b = -1
b = -2

Radius
= √[(1 - 2)2 + (-2 + 1)2]
= √2

Equation of the circle:
(x - 1)2 + (y + 2)2 = (√2)2
x2 - 2x + 1 + y2 + 4y + 4 = 2
x2 + y2 - 2x + 4y + 3 = 0

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The centres of circle C and circle C1: x2 +y2 - 12x +6y -3 =0 are identical. IF the circle C passes through the point (1,2), find its equation.

Centre of C1 = (12/2, -6/2) = (6, -3)
Centre of C = (6, -3)

Radius of C
= √[(6 - 1)2 + (-3 - 2)2]
= √50

Equation of the circle:
(x - 6)2 + (y + 3)2 = (√50)2
x2 - 12x + 36 + y2 + 6y + 9 = 50
x2 + y2 - 12x + 6y - 5 = 0
=