Question

Let the tangent to the circle C₁: x² + y² = 2 at the point M(–1, 1) intersect the circle C₂: (x – 3)² + (y – 2)² = 5, at two distinct points A and B. If the tangents to C₂ at the points A and B intersect at N, then the area of the triangle ANB is equal to

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

Answer 1

The Correct Option is C

 To find the area of triangle \(ANB\), we follow these steps:

1. Equation of Tangent to \(C_1\): The circle is \(x^2 + y^2 = 2\), centered at the origin with radius \(\sqrt{2}\). The tangent to a circle at a point \((x_1, y_1)\) is given by: \(xx_1 + yy_1 = r^2\). 
   Substituting \(M(-1, 1)\) and \(r^2 = 2\), the equation becomes: \(-x + y = 2\).
2. Intersection with Circle \(C_2\): The equation of circle \(C_2\) is: \((x - 3)^2 + (y - 2)^2 = 5\). 
   Substitute \(y = x + 2\) in the circle's equation: \((x - 3)^2 + ((x + 2) - 2)^2 = 5\). 
   Simplify to find \(x\): \((x - 3)^2 + x^2 = 5\). 
   Expanding and re-arranging gives: \(2x^2 - 6x + 4 = 0\). 
   Divide by 2: \(x^2 - 3x + 2 = 0\). 
   Solving yields roots \(x = 1\) and \(x = 2\). 
   For \(x = 1\), \(y = 1 + 2 = 3\); point \(A(1, 3)\). 
   For \(x = 2\), \(y = 2 + 2 = 4\); point \(B(2, 4)\).
3. Tangents at Points \(A\) and \(B\): The tangents at these points are perpendicular to the radius: 
   At \(A(1, 3)\), the gradient with center \((3, 2)\) is \(\frac{3-2}{1-3} = \frac{1}{-2}\). 
   Perpendicular gradient = 2; equation: \(y - 3 = 2(x - 1)\) or \(y = 2x + 1\). 
   At \(B(2, 4)\), the gradient is \(\frac{4-2}{2-3} = -2\). 
   Perpendicular gradient = \(\frac{1}{2}\); equation: \(y - 4 = \frac{1}{2}(x - 2)\) or \(y = \frac{1}{2}x + 3\).\)
4. Intersection \(N\) of Tangents: Solve the equations for \(y = 2x + 1\) and \(y = \frac{1}{2}x + 3\). 
   Equating, \(2x + 1 = \frac{1}{2}x + 3\): 
   Rearrange to get: \(4x - x = 6 - 2\), \(3x = 4\). 
   So, \(x = \frac{4}{3}\). Substituting in \(y = 2x + 1\) gives: \(y = \frac{8}{3} + 1 = \frac{11}{3}\). 
   Thus, \(N\left(\frac{4}{3}, \frac{11}{3}\right)\).
5. Area of Triangle \(ANB\): Using the formula \(\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\), with \(A(1, 3)\), \(B(2, 4)\), \(N\left(\frac{4}{3}, \frac{11}{3}\right)\): 
   \(\text{Area} = \frac{1}{2} \left| 1(4 - \frac{11}{3}) + 2(\frac{11}{3} - 3) + \frac{4}{3}(3 - 4) \right|\). 
   Simplify: \(\frac{1}{2} \left| 1(\frac{-1}{3}) + 2(\frac{2}{3}) + \frac{4}{3}(-1) \right|\). 
   \(\text{Area} = \frac{1}{2} \left| -\frac{1}{3} + \frac{4}{3} - \frac{4}{3} \right| = \frac{1}{2} \left| -\frac{1}{3} \right| = \frac{1}{6}\).

The area of triangle \(ANB\) is \(\frac{1}{6}\).