Mathematics - Circle Question with Solution | TestHub

${\mathrm{m}}_1{\mathrm{m}}_2=-1$ so right angle equation circle is
$\begin{array}{rl}& (x-4)(x-0)+(y-6)(y-0)=0\\ & x^2+y^2-4x-6y=0\end{array}$
$(\mathrm{k},3\mathrm{k})$ lies on it so
$\begin{array}{rl}& {\mathrm{k}}^2+9{\mathrm{k}}^2-4\mathrm{k}-18\mathrm{k}=0\\ & 10{\mathrm{k}}^2-22\mathrm{k}=0\end{array}$
$\mathrm{k}=0,\frac{11}{5}$
$\mathrm{k}=0$ is not possible so $\mathrm{k}=\frac{11}{5}$
also $\mathrm{r}=\sqrt{4+9}=\sqrt{13}$
so $10\mathrm{k}+{\mathrm{r}}^2=10\cdot \frac{11}{5}+(\sqrt{13}{)}^2=35$