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If the tangents at the points P and Q are the circle $\mathrm{x^{2}+y^{2}-2x+y=5}$ meet at the point $\mathrm{R\left ( \frac{9}{4},2 \right ),}$ then the area
of the triangle PQR is :
Option: 1
$\frac{5}{4}$

Option: 2
$\frac{13}{4}$

Option: 3
$\frac{5}{8}$

Option: 4
$\frac{13}{8}$

Answers (1)

best_answer

$\mathrm{x^{2}+y^{2}-2x+y=5}$

With resperct to R PQ is C.O.C

eqⁿ of C.O.C is $\mathrm{T=0}$

$\mathrm{ \frac{9}{4} x+2 y-\left(x+\frac{9}{4}\right)+\frac{1}{2}(y+2)-5=0 }$

$\mathrm{ \frac{5}{4} x+\frac{5}{2} y-\frac{25}{4}=0 }$
$\mathrm{5 x+10 y-25=0 }$
$\mathrm{x+2 y=5 }$
Area $\mathrm{=\frac{1}{2}\left(\mathrm{P}^{\prime}\right)(\mathrm{PQ}) \: \: \: \: \, \, \: \: \: \: \: \: \: \: \: \: \: \: (\mathrm{PQ})=2 \sqrt{\mathrm{r}^2-\mathrm{p}^2}=\sqrt{5} }$
$\mathrm{ =\frac{1}{2}\left[\frac{\sqrt{5}}{4}\right](\sqrt{5}) \: \: \: \: \: \: \: \: \: \: \: \: \: \mathrm{P}^{\prime}=\frac{\frac{9}{4}+4-5}{\sqrt{5}} }$
$\mathrm{=\frac{5}{8} \: \: \: \: \: \: \: \: =\left(\frac{5}{4 \sqrt{5}}\right)=\frac{\sqrt{5}}{4}}$
Method II

$\mathrm{area =\frac{RL^{3}}{R^{2}+L^{2}}}$

$\mathrm{R=\frac{5}{2}}$

$\mathrm{L=\sqrt{\frac{81}{16}+4-\frac{9}{2}+2-5}}$

$=\frac{5}{4}$

$\mathrm{area}==\frac{5}{4}$

Posted by

Ritika Kankaria

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