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Equation of Family of Circles Passing through Points of Intersection of Circle and a Line

The equation ...

Question

The equation of a circle passing through $(1, 1)$ and 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 - 25 = 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} - 17 x - 10 y + 25 = 0

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Solution

The correct option is A $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$
The equation of the required circle is $x^{2} + y^{2} + 13 x - 3 y + λ (x^{2} + y^{2} + 2 x - \frac{7}{2} y - \frac{25}{2}) = 0$
This passes through (1, 1), therefore $12 + λ (- 12) = 0$ or $λ = 1$ ,
Therefore required circle is $x^{2} + y^{2} + 13 y - 3 y + x^{2} + y^{2} + 2 x - \frac{7}{2} y - \frac{25}{2} = 0$
or $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$

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Similar questions

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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