Find the equation of the circle through the points of intersection of the circles $x^{2} + y^{2} + 2 x + 3 y - 7 = 0$ and $x^{2} + y^{2} + 3 x - 2 y - 1 = 0$ and passing through the point $(1, 2)$

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Solution

$S_{1} = x^{2} + y^{2} + 2 x + 3 y - 7 = 0$
$S_{2} = x^{2} + y^{2} + 3 x - 2 y - 1 = 0$
eqn of circle passing intersection of
$S_{1}$ & $S_{2}$
$(x^{2} + y^{2} + 2 x + 3 y - 7) + λ (x^{2} + y^{2} + 3 x - 2 y - 1)$
For real parometer $λ$
$∵$ it passes through $(1, 2)$
$(1 + 4 + 2 + 6 - 7) + λ (1 + 4 + 3 - 4 - 1) = 0$
$6 + λ (3) = 0$
$λ = - 2$
So, eqn of circle
$= (x^{2} + y^{2} + 2 x + 3 y - 7) - 2 (x^{2} + y^{2} + 3 x - 2 y - 1) = 0$
$= - x^{2} - y^{2} - 4 x + 7 y - 5 = 0$
$= x^{2} + y^{2} + 4 x - 7 y + 5 = 0$

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