(a) Show that the line 4y = x - 3 touches the circle
x² + y² - 4x - 8y + 3 = 0
(b) The straight line x = 3 intersects the circle
x² + y² - 8x - 10y - 9 = 0 at the points P and Q.
Find the equations of tangents at P and Q.
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Answer:
(a) 4y = x - 3
x = 4y + 3
Sub x = 4y + 3 into x² + y² - 4x - 8y + 3 = 0
(4y + 3)² + y² - 4(4y + 3) - 8y + 3 = 0
16y² + 24y + 9 + y² - 16y - 12 - 8y + 3 = 0
17y² = 0
y = 0
Hence, x = 3
The line 4y = x - 3 touches the circle x² + y² - 4x - 8y + 3 = 0 at (3, 0)
(shown)
(b) When x = 3,
3² + y² - 8(3) - 10y - 9 = 0
y² - 10y - 24 = 0
y = -2 or y = 12
From equation, centre of circle = (4, 5)
Point P (3, -2)
Gradient between P and centre of circle = (5 - [-2]) / (4 - 3) = 7
Thus, gradient of tangent = -1/7
y + 2 = (-1/7)(x - 3)
7y + 14 = -x + 3
7y = -x - 11
Point Q (3, 12)
(I'm going to do a short cut here)
Basically, if we reflect any straight-line graph across a horizontal line, the gradient of it reverses.
So gradient of tangent at P is -1/7
Gradient of tangent at Q is 1/7.
Thus,
y - 12 = (1/7)(x - 3)
7y - 84 = x - 3
7y = x + 81
Alternative
Use implicit differentiation on the curve at x = 3 to find the gradient.
