Let the locus of the point of intersection of the perpendicular tangents drawn to the circle $x^2+y^2+6 x$ $-4 y-12=0$ be the circle $S$. Then the equation of the tangent drawn to $\mathrm{S}$ which is perpendicular to the line $6 \mathrm{x}-4 \mathrm{y}+\mathrm{k}=0$ is

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Aicharya · Accepted Answer

Given circle is $x^2+y^2+6 x-4 y-12=0$ Centre $=(-3,2)$; radius $=5$ $\therefore$ Centre of circle $(\mathrm{S})=(-3,2)$ Since $\angle A P B=90^{\circ}$, therefore $O A P B$ is square and $O A=5$ $\therefore$ Radius of circle $(S)=O P=5 \sqrt{2}$ Now, equation of circle $\mathrm{S}$ is $(x+3)^2+(y-2)^2=50 \ldots$ (i) For slope of tangent to $S$ : $2(x+3)+2(y-2) \frac{d y}{d x}=0 \Rightarrow m_1=-\frac{(x+3)}{y-2}$ Slope of given lines is $\left(m_2\right)=\frac{3}{2}$ Since both are perpendicular $\begin{aligned} & \therefore \frac{-(x+3)}{y-2} \times \frac{3}{2}=-1 \Rightarrow x+3=\frac{2}{3}(y-2) \\ & \because \frac{4}{9}(y-2)^2+(y-2)^2=50 \text { [From equation (i)] } \\ & \Rightarrow y= \pm \frac{15 \sqrt{2}}{13}+2 \Rightarrow x= \pm \frac{30 \sqrt{2}}{39}-3\end{aligned}$ Now equation of tangent is $\begin{aligned} & \Rightarrow \mathrm{y} \mp \frac{15 \sqrt{2}}{13}-2=-\frac{2}{3}\left[x \mp \frac{30 \sqrt{2}}{39}+3\right] \\ & \Rightarrow 2 x+3 y \pm 5 \sqrt{26}=0\end{aligned}$

Let the locus of the point of intersection of the perpendicular tangents drawn to the circle $x^2+y^2+6 x$…

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