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 points $(1, 2)$

x^{2} + y^{2} + 4 x - 7 y + 5 = 0

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x^{2} + y^{2} - 4 x + 7 y + 5 = 0

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x^{2} + y^{2} + \frac{8}{3} x - \frac{1}{3} y - 3 = 0

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x^{2} + y^{2} - \frac{8}{3} x + \frac{1}{3} y - 3 = 0

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Solution

The correct option is A $x^{2} + y^{2} + 4 x - 7 y + 5 = 0$
Let $P_{1} = x^{2} + y^{2} + 2 x + 3 y - 7 = 0$ .......(1)

$P_{2} = x^{2} + y^{2} + 3 x - 2 y - 1 = 0$ .......(2)

the equation of the circle through the points of intersection of the circles $x^{2} + y^{2} + 2 x + 3 y - 7 = 0$ and $x^{2} + y^{2} + 3 x - 2 y - 1 = 0$ is

$P_{1} + k P_{2} = 0$

$x^{2} + y^{2} + 2 x + 3 y - 7 + k (x^{2} + y^{2} + 3 x - 2 y - 1) = 0$
Since, it passes through $(1, 2)$
$\Rightarrow 1^{2} + 2^{2} + 2 + 6 - 7 + k (1^{2} + 2^{2} + 3 - 4 - 1) = 0$
$\Rightarrow k = - 2$
Therefore, equation of circle is
$x^{2} + y^{2} + 2 x + 3 y - 7 - 2 (x^{2} + y^{2} + 3 x - 2 y - 1) = 0$

$x^{2} + y^{2} + 4 x - 7 y + 5 = 0$

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Find the equation of the circle through the points of intersection of the circles x2+y2+2x+3y−7=0 and x2+y2+3x−2y−1=0 and passing through the points (1,2)

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 points $(1, 2)$