The line 2x - 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

$2 x + 6 = y$
$x^{2} + y^{2} - 2 y - 9 = 0$
$x^{2} + (2 x + 6)^{2} - 2 y - 9 = 0$
$x^{2} + 4 x^{2} + 36 + 24 x - 2 y - 9 = 0$
$5 x^{2} + 24 x + 27 - 2 (2 x + 6) = 0$
$5 x^{2} + 24 x + 27 - 4 x - 12 = 0$
$5 x^{2} + 20 x + 15 = 0$
$x^{2} + 4 x + 3 = 0$
$x^{2} + 3 x + x + 3 = 0$
$x (x + 3) + 1 (x + 3) = 0$
$(x + 3) (x + 1) = 0$
$x = - 1, - 3$
$y = 4, 0$
So. These points will be Diametrically end's points.
$(x + 1) (x + 3) + (y - 4) (y - 0) = 0$
$x^{2} + 4 x + 3 + y^{2} - 4 y = 0$
$x^{2} + y^{2} + 4 x - 4 y + 3 = 0$

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