The area of the triangle formed by the tangents from the point $(4,3)$ to the circle $x^{2}+y^{2}=9$

Question

The area of the triangle formed by the tangents from the point $(4,3)$ to the circle $x^{2}+y^{2}=9$ and the line joining their points of contact is M/N in the lowest form. Find M+N

Professor PC Thomas Classes · Answer

Let $P(4,3)$ be the given point, $P T$ and $P T^{\prime}$ be the tangents from $P$\ $(4,3)$\ \ to the circle $x^{2}+y^{2}=9$, whose centre is $O(0,0)$\ Let $R$ be the point of intersection of $O P$ and $T T^{\prime}$\ Then $O P$ is the perpendicular bisector of $T T^{\prime}$ Equation of chord of contact $T T^{\prime}$ is $4 x+3 y=9$\ $O R=$ length of the perpendicular from $O$ to $T T^{\prime}$\ $=\left|\frac{4(0)+3(0)-9}{\sqrt{16+9}}ight|=\frac{9}{5}$\ $O T=$ radius of the circle $=3$\ $	herefore T R=\sqrt{O T^{2}-O R^{2}}=\sqrt{9-\frac{81}{25}}=\frac{12}{5}$\ Again $O P=\sqrt{(4-0)^{2}+(3-0)^{2}}=5$\ $	herefore P R=O P-O R=5-\frac{9}{5}=\frac{16}{5}$\ Area of the $	riangle P T T^{\prime}=P R 	imes T R=\frac{16}{5} 	imes \frac{12}{5}=\frac{192}{25}$\ sq. units.\