If the circles x^{2}+y^{2}-2x-2y-7=0 and x^{2 | vedclass

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(A) Given circles are $S_{1} \equiv x^{2}+y^{2}-2x-2y-7=0$ and $S_{2} \equiv x^{2}+y^{2}+4x+2y+k=0$.Here,$g_{1}=-1, f_{1}=-1, c_{1}=-7$ and $g_{2}=2,

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$\frac{12}{\sqrt{13}}$

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(A) Given circles are $S_{1} \equiv x^{2}+y^{2}-2x-2y-7=0$ and $S_{2} \equiv x^{2}+y^{2}+4x+2y+k=0$.Here,$g_{1}=-1, f_{1}=-1, c_{1}=-7$ and $g_{2}=2, f_{2}=1, c_{2}=k$.Since the circles cut orthogonally,$2(g_{1}g_{2} + f_{1}f_{2}) = c_{1} + c_{2}$.$2((-1)(2) + (-1)(1)) = -7 + k$$2(-2 - 1) = -7 + k$$-6 = -7 + k \Rightarrow k = 1$.The equation of the common chord is $S_{1} - S_{2} = 0$.$(x^{2}+y^{2}-2x-2y-7) - (x^{2}+y^{2}+4x+2y+1) = 0$$-6x - 4y - 8 = 0 \Rightarrow 3x + 2y + 4 = 0$.For circle $S_{1}$,center $C_{1} = (1, 1)$ and radius $r_{1} = \sqrt{1^{2} + 1^{2} - (-7)} = \sqrt{9} = 3$.The perpendicular distance $d$ from $C_{1}(1, 1)$ to the chord $3x + 2y + 4 = 0$ is:$d = \frac{|3(1) + 2(1) + 4|}{\sqrt{3^{2} + 2^{2}}} = \frac{|9|}{\sqrt{13}} = \frac{9}{\sqrt{13}}$.The length of the common chord is $2\sqrt{r_{1}^{2} - d^{2}}$.$= 2\sqrt{3^{2} - (\frac{9}{\sqrt{13}})^{2}} = 2\sqrt{9 - \frac{81}{13}} = 2\sqrt{\frac{117 - 81}{13}} = 2\sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}}$.

If the circles $x^{2}+y^{2}-2x-2y-7=0$ and $x^{2}+y^{2}+4x+2y+k=0$ cut orthogonally,then the length of the common chord of the circles is

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