JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
Two tangents are drawn from the point \(\mathrm{P}(-1,1)\) to the circle \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{x}-6 \mathrm{y}+6=0\). If these tangents touch the circle at points \(A\) and \(B\), and if \(D\) is a point on the circle such that length of the segments \(A B\) and \(A D\) are equal, then the area of the triangle \(A B D\) is eqaul to:
- A \(2\)
- B \((3 \sqrt{2}+2)\)
- C \(4\)
- D \(3(\sqrt{2}-1)\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(\triangle \mathrm{ABD}=\frac{1}{2} \times 2 \times 4\) \(=4\)
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