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If the circles $x^2+y^2-2 x-2 y-7=0$ and $x^2+y^2+4 x+2 y+k=0$ cut orthogonally, then the length of their common chord is units
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Verified Answer
The correct answer is:
$\frac{12}{\sqrt{13}}$
Since the given circles
$$
x^2+y^2-2 x-2 y-7=0 \text { and }
$$
$x^2+y^2+4 x+2 y+k=0$ cut orthogonally, then
$$
(-2)(2)+(-2)(1)=k-7
$$
$\Rightarrow \quad-4-2=k-7 \Rightarrow k=1$
So, equation of common chord is
$$
6 x+4 y+8=0 \Rightarrow 3 x+2 y+4=0
$$
$\therefore$ Length of the chord, $A B=2(A M)$
$$
x^2+y^2-2 x-2 y-7=0
$$
$$
\begin{aligned}
& =2 \sqrt{A C^2-C M^2} \\
& =2 \sqrt{9-\frac{(3+2+4)^2}{9+4}} \\
& =2 \sqrt{\frac{117-81}{13}}=2 \times \sqrt{\frac{36}{13}}=\frac{12}{\sqrt{13}} \text { units }
\end{aligned}
$$
$$
x^2+y^2-2 x-2 y-7=0 \text { and }
$$
$x^2+y^2+4 x+2 y+k=0$ cut orthogonally, then
$$
(-2)(2)+(-2)(1)=k-7
$$
$\Rightarrow \quad-4-2=k-7 \Rightarrow k=1$
So, equation of common chord is
$$
6 x+4 y+8=0 \Rightarrow 3 x+2 y+4=0
$$
$\therefore$ Length of the chord, $A B=2(A M)$
$$
x^2+y^2-2 x-2 y-7=0
$$
$$
\begin{aligned}
& =2 \sqrt{A C^2-C M^2} \\
& =2 \sqrt{9-\frac{(3+2+4)^2}{9+4}} \\
& =2 \sqrt{\frac{117-81}{13}}=2 \times \sqrt{\frac{36}{13}}=\frac{12}{\sqrt{13}} \text { units }
\end{aligned}
$$
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