P is a point of intersection of the circles S | vedclass

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(D) Let the centres of the circles $S=0$ and $S'=0$ be $C$ and $C'$ respectively.For $S \equiv x^2+y^2-6x+2ky+1=0$,the centre $C = (3, -k)$ and radius

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$\frac{\sqrt{65}}{2}$

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(D) Let the centres of the circles $S=0$ and $S'=0$ be $C$ and $C'$ respectively.For $S \equiv x^2+y^2-6x+2ky+1=0$,the centre $C = (3, -k)$ and radius $r = \sqrt{3^2+(-k)^2-1} = \sqrt{8+k^2}$.For $S' \equiv x^2+y^2+2kx-6y-7=0$,the centre $C' = (-k, 3)$ and radius $r' = \sqrt{(-k)^2+3^2-(-7)} = \sqrt{k^2+16}$.Since the tangent at $P$ to $S=0$ passes through $C'$,$CP \perp C'P$. Similarly,the tangent at $P$ to $S'=0$ passes through $C$,so $C'P \perp CP$.Thus,$	riangle CPC'$ is a right-angled triangle at $P$,where $CP = r$ and $C'P = r'$.By Pythagoras theorem,$CP^2 + C'P^2 = CC'^2$.$r^2 + r'^2 = (3 - (-k))^2 + (-k - 3)^2$.$(8+k^2) + (k^2+16) = (3+k)^2 + (-(k+3))^2$.$2k^2 + 24 = 2(k+3)^2 = 2(k^2+6k+9) = 2k^2+12k+18$.$24 = 12k + 18$ $\Rightarrow 12k = 6$ $\Rightarrow k = \frac{1}{2}$.The radius of $S'=0$ is $r' = \sqrt{k^2+16} = \sqrt{(\frac{1}{2})^2+16} = \sqrt{\frac{1}{4}+16} = \sqrt{\frac{65}{4}} = \frac{\sqrt{65}}{2}$.

$P$ is a point of intersection of the circles $S \equiv x^2+y^2-6x+2ky+1=0$ and $S' \equiv x^2+y^2+2kx-6y-7=0$. If the tangent at $P$ to $S=0$ passes through the centre of $S'=0$ and the tangent at $P$ to $S'=0$ passes through the centre of $S=0$,then the radius of $S'=0$ is

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