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

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(D) Given circles are $S_1: x^2+y^2-2x-2y-7=0$ and $S_2: x^2+y^2+4x+2y+k=0$.Since they cut orthogonally,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1=-1,

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

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(D) Given circles are $S_1: x^2+y^2-2x-2y-7=0$ and $S_2: x^2+y^2+4x+2y+k=0$.Since they cut orthogonally,$2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1=-1, f_1=-1, c_1=-7$ and $g_2=2, f_2=1, c_2=k$.$2(-1)(2) + 2(-1)(1) = -7 + k$ $\Rightarrow -4 - 2 = -7 + k$ $\Rightarrow k = 1$.The equation of the common chord is $S_1 - S_2 = 0$.$(x^2+y^2-2x-2y-7) - (x^2+y^2+4x+2y+1) = 0$ $\Rightarrow -6x - 4y - 8 = 0$ $\Rightarrow 3x + 2y + 4 = 0$.For circle $S_1$,center $C = (1, 1)$ and radius $r = \sqrt{1^2+1^2-(-7)} = \sqrt{9} = 3$.The length of the perpendicular from $C(1, 1)$ to the chord $3x+2y+4=0$ is $d = \frac{|3(1)+2(1)+4|}{\sqrt{3^2+2^2}} = \frac{9}{\sqrt{13}}$.The length of the common chord is $2\sqrt{r^2-d^2} = 2\sqrt{3^2 - (\frac{9}{\sqrt{13}})^2} = 2\sqrt{9 - \frac{81}{13}} = 2\sqrt{\frac{117-81}{13}} = 2\sqrt{\frac{36}{13}} = \frac{12}{\sqrt{13}}$ units.

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 their common chord is units.

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