If the tangents at the points [P] and [Q] on the circle x^2+y^2-2 x+y=5 meet at the point …

Question

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

Accepted Answer

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

On comparing with $x^2+y^2+2 g x+2 f y+c=0$

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

$\therefore$ Radius of the circle

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

Length of tangent, $P R=\sqrt{S_1}$

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

$\begin{aligned} & \therefore \text { Area of } \triangle P Q R=\frac{r l^3}{r^2+l^2} \\ &= \frac{\left(\frac{5}{2}\right)\left(\frac{5}{4}\right)^3}{\left(\frac{5}{2}\right)^2+\left(\frac{5}{4}\right)^2}=\frac{\frac{5}{2} \times \frac{125}{64}}{\frac{25}{4}+\frac{25}{16}} \\ &=\frac{\frac{5}{2} \times \frac{125}{64} \times 16}{125}=\frac{5}{8}\end{aligned}$

Solutions · 1