The circle $x^{2} + y^{2} + 2 x + 3 y + 1 = 0$ cuts the circle $x^{2} + y^{2} + 9 x + 3 y + 2 = 0$ at $A$ and $B$ . Then find the equation of circle with $A B$ as diameter.

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Solution

$x^{2} + y^{2} + 2 x + 3 y + 1 = 0$
cuts the circle
$x^{2} + y^{2} + 2 x + 3 y + 2 = 0$
Equation of common chord $S_{1} - S_{2} = 0 (x^{2} + y^{2} + 2 x + 3 y + 1) - (x^{2} + y^{2} + 2 x + 3 y + 2) = 0 - 7 x - 1 = 0 7 x + 1 = 0$
Considering line to be the diameter
$S_{1} + λ L = 0 x^{2} + y^{2} + 2 x + 3 y + 1 + λ (7 x + 1) = 0 x^{2} + y^{2} + 2 x + 3 y + 1 + 7 x λ + λ = 0 x^{2} + y^{2} + x (2 + 7 x) + 3 y + λ + 1 = 0 - g = - (2 + 7 x) f = - 3$
As if lies an $(7 x + 1) = 0,$ we get
$- 2 - 7 x + 1 = 0 λ = \frac{- 1}{7}$
Equation of required circle
$x^{2} + y^{2} - x + 3 y + \frac{6}{7} = 0$

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