JEE Mains · Maths · STD 11 - 10.1 circle and system of circle
If the tangents drawn at the point \(O (0,0)\) and \(P (1+\sqrt{5}, 2)\) on the circle \(x ^{2}+ y ^{2}-2 x -4 y =0\) intersect at the point \(Q\), then the area of the triangle \(OPQ\) is equal to
- A \(\frac{3+\sqrt{5}}{2}\)
- B \(\frac{4+2 \sqrt{5}}{2}\)
- C \(\frac{5+3 \sqrt{5}}{2}\)
- D \(\frac{7+3 \sqrt{5}}{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{5+3 \sqrt{5}}{2}\)
Step-by-step Solution
Detailed explanation
Tangent at \(O\) \(-( x +0)-2( y +0)=0\) \(\Rightarrow x +2 y =0\) Tangent at \(P\) \(x (1+\sqrt{5})+ y \cdot 2-( x +1+\sqrt{5})-2( y +2=0)\) Put \(x=-2 y\) \(-2 y(1+\sqrt{5})+2 y+2 y-1-\sqrt{5}-2 y-4=0\)…
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