If tangents drawn at point O (0,0) and P (1+√{5}, 2) on circle x ^{2}+ y ^{2}-2 x -4 y =0 intersec

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10-1.Circle and System of Circles

hard

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

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

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

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

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

(JEE MAIN-2022)

Solution

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 $x=-2 y$

$-2 y(1+\sqrt{5})+2 y+2 y-1-\sqrt{5}-2 y-4=0$

$-2 \sqrt{5}\, y=5+\sqrt{5} \Rightarrow y=\left(\frac{\sqrt{5}+1}{2}\right)$

$Q \left(\sqrt{5}+1,-\frac{\sqrt{5}+1}{2}\right)$

Length of tangent $OQ =\frac{5+\sqrt{5}}{2}$

$\text { Area }=\frac{ RL ^{3}}{ R ^{2}+ L ^{2}}$

$R =\sqrt{5}$

$=\frac{\sqrt{5} \times\left(\frac{5+\sqrt{5}}{2}\right)^{3}}{5+\left(\frac{5+\sqrt{5}}{2}\right)^{2}}$

$=\frac{\sqrt{5}}{2} \times \frac{4 \times(125+75+75 \sqrt{5}+5 \sqrt{5})}{(20+25+10 \sqrt{5}+5)}$

$=\frac{5+3 \sqrt{5}}{2}$

Standard 11

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