MHT CET · Maths · Circle
The locus of point of intersection of the tangents to the circle \(x^2+y^2=16\), such that the angle between them is \(60^{\circ}\), is
- A \(x^2+y^2=4\)
- B \(x^2+y^2=64\)
- C \(x^2+y^2=32\)
- D \(x^2+y^2=48\)
Answer & Solution
Correct Answer
(B) \(x^2+y^2=64\)
Step-by-step Solution
Detailed explanation
\(x^2+y^2 = r^2 \csc^2\left(\frac{\theta}{2}\right)\) \(x^2+y^2 = (4)^2 \csc^2\left(\frac{60^{\circ}}{2}\right)\)
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