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…
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
- 2
- 5
- $\frac{6}{\sqrt{13}}$
- $\frac{12}{\sqrt{13}}$
Solution
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}
$
Asked in: AP EAMCET 2020 (22 Sep Shift 1)
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