JEE Main 2025 — Circle Question with Solution

${\mathrm{m}}_1{\mathrm{m}}_2=-1$ so right angle equation circle is
$\begin{aligned}
& (x-4)(x-0)+(y-6)(y-0)=0 \\ & x^2+y^2-4 x-6 y=0
\end{aligned}$<br/>$(\mathrm{k}, 3 \mathrm{k})$liesonitso<br/>$\begin{aligned}
& \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-18 \mathrm{k}=0 \\ & 10 \mathrm{k}^2-22 \mathrm{k}=0
\end{aligned}$<br/>$\mathrm{k}=0, \frac{11}{5}$<br/>$\mathrm{k}=0$isnotpossibleso$\mathrm{k}=\frac{11}{5}$<br/>also$\mathrm{r}=\sqrt{4+9}=\sqrt{13}$<br/>so$10 \mathrm{k}+\mathrm{r}^2=10 \cdot \frac{11}{5}+(\sqrt{13})^2=35$