WBJEE · Maths · Circle
If one of the diameter of the circle, given by the equation \(x^{2}+y^{2}+4 x+6 y-12=0,\) is a chord of a circle \(S,\) whose centre is \((2,-3\) ), the radius of \(S\) is
- A \(\sqrt{41}\) unit
- B \(3 \sqrt{5}\) unit
- C \(5 \sqrt{2}\) unit
- D \(2 \sqrt{5}\) unit
Answer & Solution
Correct Answer
(A) \(\sqrt{41}\) unit
Step-by-step Solution
Detailed explanation
Given, equation of circle is \(x^{2}+y^{2}+4 x+6 y-12=0\) Centre \(C(-2,-3)\) and radius \(=\sqrt{(-2)^{2}+(-3)^{2}+12}=5\) \[ x^{2}+y^{2}+4 x+6 y-12=0 \] \[ \begin{aligned} C_{1} C_{2} &=\sqrt{\left(2+{2})^{2}+(-3+3)^{2}\right.} \\ &=\sqrt{(4)^{2}+(0)^{2}}=4 \end{aligned} \]…
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