The circle through $(- 2, 5)$ , $(0, 0)$ and intersecting the circle $x^{2} + y^{2} - 4 x + 3 y - 2 = 0$ orthogonally is

2 x^{2} + 2 y^{2} - 11 x - 16 y = 0

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

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

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x^{2} + y^{2} + 2 x - 5 y + 1 = 0

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Solution

The correct option is A $2 x^{2} + 2 y^{2} - 11 x - 16 y = 0$
Let $x^{2} + y^{2} + 29 x + 2 f y = 0$ is a equation of circle which passes through $(- 2, 5)$ and $(0, 0)$ and intersecting circle
$x^{2} + y^{2} - 4 x + 3 y - 1 = 0$ orthogonal.
$∴ \sqrt{(- g + 2)^{2} + (- f - 5)^{2}} = \sqrt{g^{2} + f^{2}}$
$∴ g^{2} + f^{2} - 4 g + 4 + 10 f + 25 = g^{2} + f^{2}$
$10 f - 4 g + 29 = 0$ ---------(1)
and By using property of orthogonality of circles,
$∴ 2 g_{1} g_{2} + 2 f_{1} f_{2} = c_{1} + c_{1} + c_{2} .$
$∴ 2 x (g_{1}) (- 2) + 2 f_{1} (\frac{+ 3}{2}) = 0 \frac{- 1}{2}$
$- 4 g + 3 f = - 1$ ---------(2)
from (1) & (2),
$7 f + 28 = 0$
$f = - 4$
and $g = \frac{- 11}{4}$
So, $e q^{4}$ of circle is
$\Rightarrow x^{2} + y^{2} - \frac{- 11}{2} x - 8 y = 0$
$\Rightarrow 2 (x^{2} + y^{2}) - 11 x - 16 y = 0$

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The circle through (−2,5) , (0,0) and intersecting the circle x2+y2−4x+3y−2=0 orthogonally is

The circle through $(- 2, 5)$ , $(0, 0)$ and intersecting the circle $x^{2} + y^{2} - 4 x + 3 y - 2 = 0$ orthogonally is