An equation of the circle through $(1, 1)$ and the points of intersection of $x^{2} + y^{2} + 13 x - 3 y = 0$ and $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ is

4 x^{2} + 4 y^{2} - 30 x - 10 y - 32 = 0

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4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0

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4 x^{2} + 4 y^{2} - 43 x + 10 y + 25 = 0

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None of these

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Solution

The correct option is B $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$
Equation of circle passing through points of intersection of $x^{2} + y^{2} + 13 x - 3 y = 0$ and $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ is $2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 + k (x^{2} + y^{2} + 13 x - 3 y) = 0$
Since, it passes through $(1, 1)$
Then, $2 + 2 + 4 - 7 - 25 + k (1 + 1 + 13 - 3) = 0$
$\Rightarrow k = 2$
Therefore, equation of circle is $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$
Ans: B

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The equation of a circle passing through points of intersection of the circles $x^{2} + y^{2} + 13 x - 3 y = 0$ and

$2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0$ and point (1, 1) is

x^{2} + y^{2} + 13 x - 3 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

(1, 1)

x^{2} + y^{2} + 13 x - 13 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0

(1, 1)

x^{2} + y^{2} + 13 x - 3 y = 0

2 x^{2} + 2 y^{2} + 4 x - 7 y - 25 = 0

Find the equation of pair of tangents drawn to the circle $x^{2} + y^{2} - 4 x + 4 y = 0$ from the point (2, 2)

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