JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Consider a circle \(C\) which touches the \(y\)-axis at \((0,6)\) and cuts off an intercept \(6 \sqrt{5}\) on the \(x\) axis. Then the radius of the circle \(C\) is equal to:
- A \(\sqrt{82}\)
- B \(9\)
- C \(8\)
- D \(\sqrt{53}\)
Answer & Solution
Correct Answer
(B) \(9\)
Step-by-step Solution
Detailed explanation
\(r=\sqrt{6^{2}+(3 \sqrt{5})^{2}}\) \(=\sqrt{36+45}=9\)
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