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If the centre of a circle which passing through the points of intersection of the circles ${x^2} + {y^2} - 6x + 2y + 4 = 0$and ${x^2} + {y^2} + 2x - 4y - 6 = 0$ is on the line $y = x$, then the equation of the circle is
$7{x^2} + 7{y^2} - 10x + 10y - 11 = 0$
$7{x^2} + 7{y^2} + 10x - 10y - 12 = 0$
$7{x^2} + 7{y^2} - 10x - 10y - 12 = 0$
$7{x^2} + 7{y^2} - 10x - 12 = 0$
Solution
(c) Family of circles through points of intersection of two circles is ${S_1} + \lambda {S_2},\,\,(\lambda \ne – 1)$.
${x^2} + {y^2} – 6x + 2y + 4 + \lambda ({x^2} + {y^2} + 2x – 4y – 6) = 0$
Centre is $(3 – \lambda ,\; – 1 + 2\lambda )$.
It lies on $y = x$.
Therefore, $ – 1 + 2\lambda = 3 – \lambda $
$\Rightarrow \lambda = \frac{4}{3}$
Hence equation of circle can be found by substituting $\lambda $ in the family of circles above.
