Equation of a circle passing through points of intersection of circles {x^2} + {y^2} + 13x - 3y = 0

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10-1.Circle and System of Circles

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

A

$4{x^2} + 4{y^2} - 30x - 10y - 25 = 0$

B

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

C

$4{x^2} + 4{y^2} - 17x - 10y + 25 = 0$

D

None of these

(IIT-1983)

Solution

(b) Required equation is

$({x^2} + {y^2} + 13x – 3y) + \lambda (2{x^2} + 2{y^2} + 4x – 7y – 25) = 0$

which passes through $(1, 1)$

so $\lambda = \frac{1}{2}$.

Hence required equation is

$4{x^2} + 4{y^2} + 30x – 13y – 25 = 0$.

Standard 11

Mathematics

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