The line $2 x - y + 6 = 0$ meets the circle $x^{2} + y^{2} - 2 y - 9 = 0$ at A and B. Find the equation of the circle on AB as diameter.

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Solution

The given equation of line and circle in
$2 x - y + 6 = 0 \dots (i) x^{2} + y^{2} - 2 y — - 9 = 0 \dots (i i)$
The point of intersection of (i) and (ii) is
$x^{2} + (2 x + 6)^{2} - 2 (2 x + 6) - 9 = 0 \Rightarrow x^{2} + 4 x^{2} + 24 x + 36 - 4 x - 12 - 9 = 0 \Rightarrow 5 x^{2} + 20 x + 15 = 0 \Rightarrow x^{2} + 4 x + 3 = 0 \Rightarrow (x + 3) (x + 1) = 0 \Rightarrow x = (- 3, — 1) ∴ y = (0, 4)$
$∴$ A (-3, 0) and 13 a (-1, 4)
$∵$ AB is a diameter, so the equation of circle is
$(x + 3) (x + 1) + (y - 0) (y - 4) \Rightarrow x^{2} + y^{2} + 4 x - 4 y + 3 = 0$

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