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Equation of Family of circles passing through point of intersection of two circles

The equation ...

Question

The equation of the circle passing 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

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

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

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x^{2} + y^{2} + 15 x - 13 y - 25 = 0

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

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Solution

The correct option is B $4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$
Circle through points of intersection of given two circles is
$(x^{2} + y^{2} + 13 x - 3 y) + λ (2 x^{2} + 2 y^{2} + 4 x - 7 y - 25) = 0 \Rightarrow (1 + 2 λ) x^{2} + (1 + 2 λ) y^{2} + (13 + 4 λ) x + (- 3 - 7 λ) y - 25 λ = 0$
As it passes through (1, 1),
$∴ 1 + 2 λ + 1 + 2 λ + 13 + 4 λ - 3 - 7 λ - 25 λ = 0 \Rightarrow - 24 λ + 12 = 0 \Rightarrow λ = \frac{1}{2} ∴ R e q u i r e d c i r c l e i s 2 x^{2} + 2 y^{2} + 15 x - \frac{13 y}{2} - \frac{25}{2} = 0 o r 4 x^{2} + 4 y^{2} + 30 x - 13 y - 25 = 0$

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Similar questions

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

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

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