TS EAMCET · Maths · Circle
The length of common chord of the circles \(x^2+y^2-6 x-4 y+13-c^2=0\) and \(x^2+y^2-4 x-6 y+13-c^2=0\) is
- A \(\sqrt{4 c^2-2}\)
- B \(\frac{1}{2} \sqrt{4 c^2-2}\)
- C \(\sqrt{c^2-2}\)
- D \(\sqrt{4 c^2-1}\)
Answer & Solution
Correct Answer
(A) \(\sqrt{4 c^2-2}\)
Step-by-step Solution
Detailed explanation
Given circles, \[ \begin{aligned} & x^2+y^2-6 x-4 y+13-c^2=0 \text { and } \\ & x^2+y^2-4 x-6 y+13-c^2=0 \end{aligned} \] Centres of the circles are \(C_1(3,2)\) and \(C_2(2,3)\) and \(r_1=c\) and \(r_2=c\) The circles intersects at two points. Radical axis is the common chord…
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