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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 $S_1 + \lambda S_2 = 0$.He

Vedclass · Accepted Answer

$x^2 + y^2 - 3x + 1 = 0$

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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 $S_1 + \lambda S_2 = 0$.Here,$S_1 = x^2 + y^2 - 6 = 0$ and $S_2 = x^2 + y^2 - 6x + 8 = 0$.The equation is $(x^2 + y^2 - 6) + \lambda(x^2 + y^2 - 6x + 8) = 0$.Since the circle passes through $(1, 1)$,we substitute $x = 1$ and $y = 1$ into the equation:$(1^2 + 1^2 - 6) + \lambda(1^2 + 1^2 - 6(1) + 8) = 0$$(2 - 6) + \lambda(2 - 6 + 8) = 0$$-4 + \lambda(4) = 0$$4\lambda = 4 \implies \lambda = 1$.Substituting $\lambda = 1$ back into the equation:$(x^2 + y^2 - 6) + 1(x^2 + y^2 - 6x + 8) = 0$$2x^2 + 2y^2 - 6x + 2 = 0$Dividing by $2$,we get $x^2 + y^2 - 3x + 1 = 0$.

Find the equation of the circle passing through the point $(1, 1)$ and the intersection points of the circles $x^2 + y^2 = 6$ and $x^2 + y^2 - 6x + 8 = 0$.

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