TS EAMCET · Maths · Circle
Length of the common chord of the circles \(x^2+y^2-6 x+5=\) and \(x^2+y^2+4 y-5=0\) is
- A \(\sqrt{13}\)
- B \(\frac{12}{\sqrt{13}}\)
- C \(\frac{6}{\sqrt{13}}\)
- D \(2 \sqrt{13}\)
Answer & Solution
Correct Answer
(B) \(\frac{12}{\sqrt{13}}\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2-6 x+5\) and \(x^2+y^2+4 y-5=0\) \(\begin{aligned} & \Rightarrow(x-3)^2+y^2=4 \text { and } x^2+(y+2)^2=9 \\ & \Rightarrow a=3, b=-2 \end{aligned}\) Length of common chord \(=\left|\frac{2 a b}{\sqrt{a^2+b^2}}\right|=\frac{12}{\sqrt{13}}\).
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