JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If one of the diameters of the circle \(x^{2}+y^{2}-2 \sqrt{2} x\) \(-6 \sqrt{2} y+14=0\) is a chord of the circle \((x-2 \sqrt{2})^{2}\) \(+(y-2 \sqrt{2})^{2}=r^{2}\), then the value of \(r^{2}\) is equal to
- A \(15\)
- B \(70\)
- C \(18\)
- D \(10\)
Answer & Solution
Correct Answer
(D) \(10\)
Step-by-step Solution
Detailed explanation
\(S: x^{2}+y^{2}-2 \sqrt{2} x-6 \sqrt{2} y+14=0\) \(C(\sqrt{2}, 3 \sqrt{2}), O(2 \sqrt{2}, 2 \sqrt{2})\) \(r_{1}=\sqrt{6}\) \(S_{1}:(x-2 \sqrt{2})^{2}+(y-2 \sqrt{2})^{2}=r^{2}\) \(\text { Now in } \Delta OCQ\) \(|O C|^{2}+|C Q|^{2}=|O Q|^{2}\) \(4+6=r^{2}\) \(r^{2}=10\)
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