If S equiv 2x^2+2y^2-8x+8y-7=0 is the circle | vedclass

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(A) The equation of a circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda(S_2 - S_1) = 0$.Given $S_1: x^2+y^2+kx-ky+

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$\frac{3}{\sqrt{2}}$

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(A) The equation of a circle passing through the intersection of $S_1$ and $S_2$ is given by $S_1 + \lambda(S_2 - S_1) = 0$.Given $S_1: x^2+y^2+kx-ky+1=0$ and $S_2: x^2+y^2-kx+ky-2=0$.$S_2 - S_1 = -2kx + 2ky - 3 = 0$.So,the equation is $x^2+y^2+kx-ky+1 + \lambda(-2kx+2ky-3) = 0$.$x^2+y^2 + k(1-2\lambda)x - k(1-2\lambda)y + (1-3\lambda) = 0$.Comparing this with the given circle $S: x^2+y^2-4x+4y-\frac{7}{2} = 0$:$k(1-2\lambda) = -4$ and $1-3\lambda = -\frac{7}{2}$.From $1-3\lambda = -\frac{7}{2}$,we get $3\lambda = \frac{9}{2} \Rightarrow \lambda = \frac{3}{2}$.Substituting $\lambda = \frac{3}{2}$ into $k(1-2\lambda) = -4$:$k(1-2(\frac{3}{2})) = -4$ $\Rightarrow k(1-3) = -4$ $\Rightarrow -2k = -4$ $\Rightarrow k = 2$.The point is $(k, k) = (2, 2)$.The circle $S$ is $x^2+y^2-4x+4y-\frac{7}{2} = 0$. The center $C$ is $(2, -2)$ and radius $r = \sqrt{2^2+(-2)^2 - (-\frac{7}{2})} = \sqrt{4+4+\frac{7}{2}} = \sqrt{\frac{23}{2}}$.The length of the tangent from $(2, 2)$ to $S$ is $\sqrt{S(2, 2)} = \sqrt{2^2+2^2-4(2)+4(2)-\frac{7}{2}} = \sqrt{4+4-8+8-\frac{7}{2}} = \sqrt{8-\frac{7}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.

If $S \equiv 2x^2+2y^2-8x+8y-7=0$ is the circle passing through the points of intersection of the circles $x^2+y^2+kx-ky+1=0$ and $x^2+y^2-kx+ky-2=0$,then the length of the tangent drawn from the point $(k, k)$ to the circle $S$ is

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