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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

  1. A \(x^2+y^2=4\)
  2. B \(x^2+y^2=64\)
  3. C \(x^2+y^2=32\)
  4. D \(x^2+y^2=48\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(x^2+y^2=64\)

Step-by-step Solution

Detailed explanation

\(x^2+y^2 = r^2 \csc^2\left(\frac{\theta}{2}\right)\) \(x^2+y^2 = (4)^2 \csc^2\left(\frac{60^{\circ}}{2}\right)\)