The line x-2=0 cuts the circle x^2+y^2-8x-2y+ | vedclass

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(B) The family of circles passing through the intersection of the line $x-2=0$ and the circle $x^2+y^2-8x-2y+8=0$ is given by:$(x^2+y^2-8x-2y+8) + \la

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$x^2+y^2-4x-2y=0$

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(B) The family of circles passing through the intersection of the line $x-2=0$ and the circle $x^2+y^2-8x-2y+8=0$ is given by:$(x^2+y^2-8x-2y+8) + \lambda(x-2) = 0$$x^2+y^2+(\lambda-8)x-2y+(8-2\lambda) = 0$ ... $(i)$The center of this circle is $(-\frac{\lambda-8}{2}, 1)$.For the circle to have the least radius,the chord $AB$ must be the diameter of the circle. This means the center of the circle must lie on the line $x-2=0$.Substituting the x-coordinate of the center into the line equation:$-\frac{\lambda-8}{2} = 2$$-\lambda+8 = 4$$\lambda = 4$Substituting $\lambda=4$ into equation $(i)$:$x^2+y^2+(4-8)x-2y+(8-2(4)) = 0$$x^2+y^2-4x-2y+0 = 0$$x^2+y^2-4x-2y=0$

The line $x-2=0$ cuts the circle $x^2+y^2-8x-2y+8=0$ at $A$ and $B$. The equation of the circle passing through the points $A$ and $B$ and having the least radius is

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