TS EAMCET · Maths · Circle
\((a, 0)\) and \((b, 0)\) are centres of two circles belonging to a coaxial system of which \(y\)-axis is the radical axis. If radius of one of the circles is ' \(r\) ', then the radius of the other circle is
- A \(\left(r^2+b^2+a^2\right)^{1 / 2}\)
- B \(\left(r^2+b^2-a^2\right)^{1 / 2}\)
- C \(\left(r^2+b^2-a^2\right)^{1 / 3}\)
- D \(\left(r^2+b^2+a^2\right)^{1 / 3}\)
Answer & Solution
Correct Answer
(B) \(\left(r^2+b^2-a^2\right)^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Let the equation of circle whose centre \((a, 0)\) and radius \((r)\) is \[ \begin{gathered} (x-a)^2+(y-0)^2=r^2 \\ \Rightarrow \quad S_1 \equiv x^2+a^2-2 a x+y^2-r^2=0 \end{gathered} \] and the equation of circle whose centre \((b, 0)\) and radius \(R\) is…
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