急!F5 Equations of Circles 8q12

(a)
L : x + y = 4 ...... [1]
C : x² + y² + 4x - 2y - 20 = 0 ......[2]

[1] :
x = 4 - y ...... [3]

Put [3] into [2] :
(4 - y)² + y² + 4(4 - y) - 2y - 20 = 0
16 - 8y + y² + y² + 16 - 4y - 2y - 20 = 0
2y² - 14y + 12 = 0
y² - 7y + 6 = 0
(y - 1)(y - 6) = 0
y = 1 or y = 6

Put y = 1 into [3] :
x = 4 - (1)
x = 3

Put y = 6 into [3] :
x = 4 - (6)
x = -2

Hence, A = (-2, 6) and B = (3, 1)


(b)(i)
Let M be the centre of C.
M = (-4/2, -(-2)/2) = (-2, 1)

The x-coordinate of M is equal to that of A.
Hence, MA // y-axis
(Slope of the tangent at A) ⊥ MA
Hence, (the tangent at A) // x-axis

The tangent at A :
y = 6
y - 6 = 0


(b)(ii)
The y-coordinate of M is equal to that of B.
Hence, MB // x-axis
(Slope of the tangent at B) ⊥ MB
Hence, (the tangent at B) // y-axis

The tangent at B :
x = 3
x - 3 = 0