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Question

(B) The family of circles passing through the intersection of the circle $x^2+y^2=9$ and the line $x+y=1$ is given by $(x^2+y^2-9) + \lambda(x+y-1) =

Vedclass · Accepted Answer

$x^2+y^2-9-(x+y-1)=0$

Vedclass · Answer

(B) The family of circles passing through the intersection of the circle $x^2+y^2=9$ and the line $x+y=1$ is given by $(x^2+y^2-9) + \lambda(x+y-1) = 0$. This simplifies to $x^2+y^2+\lambda x+\lambda y - (\lambda+9) = 0$. The center of this circle is $(-\frac{\lambda}{2}, -\frac{\lambda}{2})$. For the smallest circle,the line $x+y=1$ must be the diameter of the circle. Substituting the center into the line equation: $-\frac{\lambda}{2} - \frac{\lambda}{2} = 1$,which gives $-\lambda = 1$,so $\lambda = -1$. Substituting $\lambda = -1$ into the family equation,we get $(x^2+y^2-9) - (x+y-1) = 0$.

The equation of the smallest circle passing through the intersection of the line $x+y=1$ and the circle $x^2+y^2=9$ is

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