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JEE Mains · Maths · STD 11 - 10.1 circle and system of circle

Let the tangent to the circle \(C _{1}: x^{2}+y^{2}=2\) at the point \(M (-1,1)\) intersect the circle \(C _{2}\) : \(( x -3)^{2}+(y-2)^{2}=5\), at two distinct points \(A\) and \(B\). If the tangents to \(C _{2}\) at the points \(A\) and \(B\) intersect at \(N\), then the area of the triangle \(ANB\) is equal to

  1. A \(\frac{1}{2}\)
  2. B \(\frac{2}{3}\)
  3. C \(\frac{1}{6}\)
  4. D \(\frac{5}{3}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\frac{1}{6}\)

Step-by-step Solution

Detailed explanation

\(OP =\left|\frac{2-3+2}{\sqrt{2}}\right|\) \(OP =\frac{3}{\sqrt{2}}\) \(AP =\sqrt{ OA ^{2}- OP ^{2}}\) \(=\frac{1}{\sqrt{2}}\) \(\tan \theta=3\) \(\therefore \sin \theta=\frac{3}{\sqrt{10}}=\frac{ AP }{ AN }\) \(\Rightarrow AN =\frac{\sqrt{5}}{3}= BN\) Area of…