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(B) The equation of a family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.Gi

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of a family of circles passing through the intersection of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 + \lambda S_2 = 0$.Given $S_1: x^2 + y^2 + 13x - 3y = 0$ and $S_2: 2x^2 + 2y^2 + 4x - 7y - 25 = 0$.The required equation is $(x^2 + y^2 + 13x - 3y) + \lambda (2x^2 + 2y^2 + 4x - 7y - 25) = 0$.Since the circle passes through $(1, 1)$,we substitute $x = 1$ and $y = 1$:$(1^2 + 1^2 + 13(1) - 3(1)) + \lambda (2(1^2) + 2(1^2) + 4(1) - 7(1) - 25) = 0$$(1 + 1 + 13 - 3) + \lambda (2 + 2 + 4 - 7 - 25) = 0$$12 + \lambda (-24) = 0$$12 = 24\lambda \implies \lambda = \frac{1}{2}$.Substituting $\lambda = \frac{1}{2}$ into the family equation:$(x^2 + y^2 + 13x - 3y) + \frac{1}{2}(2x^2 + 2y^2 + 4x - 7y - 25) = 0$Multiply by $2$:$2x^2 + 2y^2 + 26x - 6y + 2x^2 + 2y^2 + 4x - 7y - 25 = 0$$4x^2 + 4y^2 + 30x - 13y - 25 = 0$.

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

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