Question

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

Answer: (C)

Tangent at $O$
$-(x+0)-2(y+0)=0$
$\Rightarrow x+2y=0$
Tangent at $P$
$x(1+\sqrt{5})+y\cdot 2-(x+1+\sqrt{5})-2(y+2=0)$
Put $x=-2y$
$-2y(1+\sqrt{5})+2y+2y-1-\sqrt{5}-2y-4=0$
$-2\sqrt{5}y=5+\sqrt{5}$
$\Rightarrow y=\left(\frac{\sqrt{5}+1}{2}\right)$
$Q\left(\sqrt{5}+1,-\frac{\sqrt{5}+1}{2}\right)$
Length of tangent $OQ=\frac{5+\sqrt{5}}{2}$
Area $=\frac{R{L}^3}{R^2+L^2}$
$R=\sqrt{5}$
$=\frac{\sqrt{5}\times {\left(\frac{5+\sqrt{5}}{2}\right)}^3}{5+{\left(\frac{5+\sqrt{5}}{2}\right)}^2}$
$=\frac{\sqrt{5}}{2}\times \frac{4\times (125+75+75\sqrt{5}+5\sqrt{5})}{(20+25+10\sqrt{5}+5)}$
$=\frac{5+3\sqrt{5}}{2}$