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AP EAMCET · Maths · Straight Lines

Let the tangents drawn from \(\mathrm{P}(-1,-1)\) to the circle \(x^2+y^2-2 x-4 y-4=0\) touch the circle at the points \(\mathrm{A}\) and \(\mathrm{B}\). Then the area of the triangle \(\mathrm{PAB}\) (in square units) is

  1. A \(\frac{24}{13}\)
  2. B \(\frac{24}{7}\)
  3. C \(\frac{8}{13}\)
  4. D \(\frac{3}{13} 4^{2 / 3}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{24}{13}\)

Step-by-step Solution

Detailed explanation

Center \(C=(1,2)\), radius \(r=\sqrt{(-1)^2+(-2)^2-(-4)} = 3\). \(PC = \sqrt{(1-(-1))^2+(2-(-1))^2} = \sqrt{2^2+3^2} = \sqrt{13}\). Length of tangent \(PA = \sqrt{PC^2-r^2} = \sqrt{(\sqrt{13})^2-3^2} = \sqrt{13-9} = \sqrt{4} = 2\). Let \(\angle CPA = \alpha\). Then…
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