JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Let \(C\) be a circle passing through the points \(A (2,-1)\) and \(B (3,4)\). The line segment \(AB\) is not a diameter of \(C\). If \(r\) is the radius of \(C\) and its centre lies on the circle \((x-5)^{2}+(y-1)^{2}=\frac{13}{2}\), then \(r^{2}\) is equal to
- A \(32\)
- B \(\frac{65}{2}\)
- C \(\frac{61}{2}\)
- D \(30\)
Answer & Solution
Correct Answer
(B) \(\frac{65}{2}\)
Step-by-step Solution
Detailed explanation
\(AB =\sqrt{26}\) \(r ^{2}= CM ^{2}+ AM ^{2}\) \(=\left(2 \times \sqrt{\frac{13}{2}}\right)^{2}+\left(\sqrt{\frac{13}{2}}\right)^{2}\) \(r ^{2}=\frac{65}{2}\)
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