TS EAMCET · Maths · Circle
The length of the common chord of the two circles \((x-a)^2+y^2=a^2\) and \(x^2+(y-b)^2=b^2\), is
- A \(\frac{a b}{\sqrt{a^2+b^2}}\)
- B \(\frac{2 a b}{\sqrt{a^2+b^2}}\)
- C \(\frac{a+b}{\sqrt{a^2+b^2}}\)
- D \(\sqrt{a^2+b^2}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 a b}{\sqrt{a^2+b^2}}\)
Step-by-step Solution
Detailed explanation
The equation of the common chord \(P Q\) of the circles \( \begin{array}{r} S_1: x^2+y^2-2 a x=0 \\ S_2: x^2+y^2-2 b y=0 \\ S_1-S_2=0 \\ 2 b y-2 a x=0 \\ a x-b y=0 \end{array} \) The centre of \(S_1\) is \((a, 0)\) and radius is a. The length of the perpendicular from \((a, 0)\)…
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