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(A) Using the family of circles,the equation of the circle passing through the points of intersection of the circles $S_1: x^2+y^2+13x-3y=0$ and $S_2:

Vedclass · Accepted Answer

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

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(A) Using the family of circles,the equation of the circle passing through the points of intersection of the circles $S_1: x^2+y^2+13x-3y=0$ and $S_2: 2x^2+2y^2+4x-7y-25=0$ is given by $S_2 + \lambda S_1 = 0$. Substituting the equations,we get: $2x^2+2y^2+4x-7y-25 + \lambda(x^2+y^2+13x-3y) = 0$. Since the circle passes through $(1,1)$,we substitute $x=1$ and $y=1$ into the equation: $2(1)^2+2(1)^2+4(1)-7(1)-25 + \lambda(1^2+1^2+13(1)-3(1)) = 0$. $2+2+4-7-25 + \lambda(1+1+13-3) = 0$. $-24 + \lambda(12) = 0$. $12\lambda = 24 \Rightarrow \lambda = 2$. Substituting $\lambda = 2$ back into the family equation: $2x^2+2y^2+4x-7y-25 + 2(x^2+y^2+13x-3y) = 0$. $2x^2+2y^2+4x-7y-25 + 2x^2+2y^2+26x-6y = 0$. $4x^2+4y^2+30x-13y-25 = 0$.

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

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