$\mathrm{P}$ is a point of intersection of the circles $\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}$ $+2 k y+1=0$ and $S^1 \equiv x^2+y^2+2 k x-6 y-7=0$. If the tangent at $P$ to $S=0$ pass through the centre of $S^1=0$ and the tangent at $\mathrm{P}$ to $\mathrm{S}^1=0$ pass through…

Question

$\mathrm{P}$ is a point of intersection of the circles $\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2-6 \mathrm{x}$ $+2 k y+1=0$ and $S^1 \equiv x^2+y^2+2 k x-6 y-7=0$. If the tangent at $P$ to $S=0$ pass through the centre of $S^1=0$ and the tangent at $\mathrm{P}$ to $\mathrm{S}^1=0$ pass through the centre of $\mathrm{S}=0$, then the radius of $\mathrm{S}^1=0$ is

Aicharya · Accepted Answer

From given equation of circles $\begin{aligned} & \mathrm{C}^{\prime}(3,-\mathrm{k}), \mathrm{r}^{\prime}=\sqrt{9+\mathrm{k}^2-1} \\ & \text { and } \mathrm{C}^{\prime}(-\mathrm{k}, 3), \mathrm{r}^{\prime}=\sqrt{9+\mathrm{k}^2+7} \end{aligned}$ According to question $\begin{aligned} & \mathrm{CP}^2+\mathrm{C}^{\prime} \mathrm{P}^2=\mathrm{CC}^{\prime}\left(\because \angle \mathrm{CPC}^{\prime}=90^{\circ}\right) \\ & \Rightarrow 18+2 \mathrm{k}^2+6=2(3+\mathrm{k})^2 \Rightarrow \mathrm{k}=\frac{1}{2} \\ & \text { Now, } \mathrm{r}^{\prime}=\sqrt{9+\mathrm{k}^2+7}=\frac{\sqrt{65}}{2} \end{aligned}$

$\mathrm{P}$ is a point of intersection of the circles $\mathrm{S} \equiv \mathrm{x}^2+\mathrm{y}^2-6…

Solution