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(A) Let $S_{1} = x^{2}+y^{2}-6x-8=0$ and $S_{2} = x^{2}+y^{2}-6=0$.The equation of the family of circles passing through the intersection of $S_{1}$ a

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$x^{2}+y^{2}+3x-5=0$

Vedclass · Answer

(A) Let $S_{1} = x^{2}+y^{2}-6x-8=0$ and $S_{2} = x^{2}+y^{2}-6=0$.The equation of the family of circles passing through the intersection of $S_{1}$ and $S_{2}$ is given by $S_{1} + \lambda S_{2} = 0$.$(x^{2}+y^{2}-6x-8) + \lambda(x^{2}+y^{2}-6) = 0 \quad \dots(i)$Since the circle passes through the point $(1, 1)$, we substitute $x=1$ and $y=1$ into equation $(i)$:$(1^{2}+1^{2}-6(1)-8) + \lambda(1^{2}+1^{2}-6) = 0$$(1+1-6-8) + \lambda(1+1-6) = 0$$-12 + \lambda(-4) = 0$$-4\lambda = 12 \Rightarrow \lambda = -3$.Substituting $\lambda = -3$ back into equation $(i)$:$(x^{2}+y^{2}-6x-8) - 3(x^{2}+y^{2}-6) = 0$$x^{2}+y^{2}-6x-8 - 3x^{2}-3y^{2}+18 = 0$$-2x^{2}-2y^{2}-6x+10 = 0$Dividing by $-2$, we get $x^{2}+y^{2}+3x-5 = 0$.

The equation of the circle passing through the point $(1, 1)$ and the points of intersection of $x^{2}+y^{2}-6x-8=0$ and $x^{2}+y^{2}-6=0$ is

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