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  • If the tangents drawn at the points <img alt="\mathrm{O\left ( 0,0 \right )\: and\: P\left ( 1+\sqrt{5},2 \right )}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BO%5Cleft%20%28%

If the tangents drawn at the points \mathrm{O\left ( 0,0 \right )\: and\: P\left ( 1+\sqrt{5},2 \right )} on the circle \mathrm{x^{2}+y^{2}-2x-4y= 0} intersect at the point \mathrm{Q}. then the area of the triangle \mathrm{OPQ} is equal to

Option: 1

\mathrm{\frac{3+\sqrt{5}}{2}}


Option: 2

\mathrm{\frac{4+2\sqrt{5}}{2}}


Option: 3

\mathrm{\frac{5+3\sqrt{5}}{2}}


Option: 4

\mathrm{\frac{7+3\sqrt{5}}{2}}


Answers (1)

best_answer

\mathrm{Tangent \: at\: (0,0)\: is \: \: T=0}
\mathrm{\Rightarrow x \cdot 0+y \cdot 0-(x+0)-2(y+0)=0}
\mathrm{\Rightarrow x+2 y=0 \, ---(1)}

\mathrm{Tangent \: at (1+\sqrt{5}, 2)\: is}
\mathrm{ x(1+\sqrt{5})+y \cdot 2-(x+1+\sqrt{5})-2(y+2)=0 \\ }
\mathrm{ \Rightarrow \sqrt{5} x=\sqrt{5}+5 \Rightarrow x=\sqrt{5}+1\: ---(2) }

Point of intersection of (1) & (2) is \mathrm{ \left(\sqrt{5}+1,\frac{-(\sqrt{5}+1)}{2}\right)

\mathrm{Area \: of \: \Delta O P Q=\left | \frac{1}{2} \right |\begin{vmatrix} 1 &0 &0 \\ 1& 1+\sqrt{5}& 2\\ 1& 1+\sqrt{5} & \frac{-\sqrt{5}+1}{2} \end{vmatrix}\mid}

\mathrm{=\frac{1}{2}(\sqrt{5}+1)\left(\frac{\sqrt{5}+1}{2}+2\right)=\frac{1}{2}\left[\frac{(\sqrt{5}+1)(\sqrt{5}+5)}{2}\right]}
\mathrm{=\frac{5+5 \sqrt{5}+\sqrt{5}+5}{4}=\frac{10+6 \sqrt{5}}{4}=\frac{5+3 \sqrt{5}}{2}}
Option (C)
 

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Gaurav

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