If the tangents at the points P and Q on the circle x^2+y^2-2x+y=5 meet at the point R((9)…

Question

If the tangents at the points $P$ and $Q$ on the circle ${x}^{2}+{y}^{2}-2x+y=5$ meet at the point $R(\frac{9}{4},2)$, then the area of the triangle $PQR$ is :

Accepted Answer

Equation of circle ${x}^{2}+{y}^{2}-2x+y-5=0$

On comparing with ${x}^{2}+{y}^{2}+2gx+2fy+c=0$

$\begin{gathered} 2g=-2,2f=1,c=-5 \ g=-1,f=\frac{1}{2},c=-5 \end{gathered}$

$	herefore $ Radius of the circle

$r=\sqrt{(-1{)}^{2}+{(\frac{1}{2})}^{2}+5}=\frac{5}{2}$

Length of tangent, $PR=\sqrt{{S}_{1}}$

$l=\sqrt{{(\frac{9}{4})}^{2}+(2{)}^{2}-2(\frac{9}{4})+2-5}=\frac{5}{4}$

$\begin{gathered} 	herefore 	rianglePQR=\frac{r{l}^{3}}{{r}^{2}+{l}^{2}} \ =\frac{(\frac{5}{2}){(\frac{5}{4})}^{3}}{{(\frac{5}{2})}^{2}+{(\frac{5}{4})}^{2}}=\frac{\frac{5}{2}	imes \frac{125}{64}}{\frac{25}{4}+\frac{25}{16}} \ =\frac{\frac{5}{2}	imes \frac{125}{64}	imes 16}{125}=\frac{5}{8} \end{gathered}$

Solutions · 1