
m1 m2=−1 so right angle equation circle is
$\begin{aligned}
& (x-4)(x-0)+(y-6)(y-0)=0 \
& x^2+y^2-4 x-6 y=0
\end{aligned}$
(k,3k) lies on it so
$\begin{aligned}
& \mathrm{k}^2+9 \mathrm{k}^2-4 \mathrm{k}-18 \mathrm{k}=0 \
& 10 \mathrm{k}^2-22 \mathrm{k}=0
\end{aligned}$
k=0,511
k=0 is not possible so k=511
also r=4+9=13
so 10k+r2=10⋅511+(13)2=35