The centre of the circle whose radius is 3 units and touching internally the circle $x^2+y^2-4 x-6 y-12=0$…
- $\left(\frac{4}{5}, \frac{7}{5}\right)$
- $\left(\frac{4}{5}, \frac{-7}{5}\right)$
- $\left(\frac{-4}{5}, \frac{-7}{5}\right)$
- $\left(\frac{-4}{5}, \frac{7}{5}\right)$
Solution
$\begin{aligned}
\mathrm{PC}_1 & =\sqrt{(2+1)^2+(3+1)^2} \\
& =\sqrt{25} \\
& =5
\end{aligned}$
$P$ divides $C_1 C_2$ externally in the ratio $r_1: r_2$ i.e. $5: 3$
$\therefore \quad-1=\frac{5(\mathrm{~h})-3(2)}{5-3}$ and $-1=\frac{5(\mathrm{k})-3(3)}{5-3}$
$\begin{aligned} & \Rightarrow-2=5 \mathrm{~h}-6 \text { and }-2=5 \mathrm{k}-9 \\ & \Rightarrow \mathrm{h}=\frac{4}{5} \text { and } \mathrm{k}=\frac{7}{5}\end{aligned}$Asked in: MHT CET 2023 (14 May Shift 2)