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Question

(A) The equation of a circle passing through the points of intersection of the two circles is given by $S_1 + \lambda S_2 = 0$,where $\lambda 
eq -1$

Vedclass · Accepted Answer

$x^2+y^2-2x-12=0$

Vedclass · Answer

(A) The equation of a circle passing through the points of intersection of the two circles is given by $S_1 + \lambda S_2 = 0$,where $\lambda 
eq -1$. $(x^2+y^2-4x-6y-12) + \lambda(x^2+y^2+6x+4y-12) = 0$ $(1+\lambda)x^2 + (1+\lambda)y^2 + (6\lambda-4)x + (4\lambda-6)y - 12(1+\lambda) = 0$ Dividing by $(1+\lambda)$,we get $x^2+y^2 + \frac{6\lambda-4}{1+\lambda}x + \frac{4\lambda-6}{1+\lambda}y - 12 = 0$. Comparing with $x^2+y^2+2gx+2fy+c=0$,we have $g = \frac{3\lambda-2}{1+\lambda}$,$f = \frac{2\lambda-3}{1+\lambda}$,and $c = -12$. The radius $r = \sqrt{g^2+f^2-c} = \sqrt{13}$. $g^2+f^2-c = 13 \implies \frac{(3\lambda-2)^2 + (2\lambda-3)^2}{(1+\lambda)^2} + 12 = 13$. $(3\lambda-2)^2 + (2\lambda-3)^2 = (1+\lambda)^2$. $9\lambda^2 - 12\lambda + 4 + 4\lambda^2 - 12\lambda + 9 = 1 + 2\lambda + \lambda^2$. $12\lambda^2 - 26\lambda + 12 = 0 \implies 6\lambda^2 - 13\lambda + 6 = 0$. $(2\lambda-3)(3\lambda-2) = 0$,so $\lambda = \frac{3}{2}$ or $\lambda = \frac{2}{3}$. For $\lambda = \frac{2}{3}$,the equation becomes $x^2+y^2-2x-12=0$. For $\lambda = \frac{3}{2}$,the equation becomes $x^2+y^2+2x-12=0$.

The equation of a circle passing through the points of intersection of the circles $x^2+y^2-4x-6y-12=0$ and $x^2+y^2+6x+4y-12=0$ and having radius $\sqrt{13}$ is

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