Given lines are $5 x+6 y-34=0$ and $2 x+y+C=0$ with respect to $x^2+y^2-8 x-10 y+25=0$.
$$
\begin{aligned}
& x^2+y^2-8 x-10 y+25+16-16=0 \\
& \left(x^2-8 x+16\right)+\left(y^2-10 y+25\right)=16 \\
& (x-4)^2+(y-5)^2=(4)^2
\end{aligned}
$$
Here, centre is $(4,5)$ and radius is 4 .
Take, $\mathrm{X}=x-4, \mathrm{Y}=y-5$
$$
x=\mathrm{X}+4, y=\mathrm{Y}+5
$$
Then, $\mathrm{X}^2+\mathrm{Y}^2=16$
Put the transition coordinate in the given lines.
$$
\begin{aligned}
& 5 x+6 y-34=0 \\
& 5(x+4)+6(y+5)-34=0 \\
& 5 x+6 y+20+30-3 y=0 \\
& 5 x+6 y+50-34=0 \\
& 5 x+6 y+16=0 \\
& 2 x+y+c=0
\end{aligned}
$$
$$
\begin{aligned}
& 2(x+4)+(y+5)+\mathrm{c}=0 \\
& 2 x+8+y+5+\mathrm{c}=0 \\
& 2 x+y+(13+\mathrm{c})=0 \\
& \text { Apply } r^2\left(l_1 l_2+m_1 m_2\right)=n_1 n_2 \\
& (16)(5 \times 2+(\text { ii }) \\
& 10+6=13+\mathrm{c} \\
& \mathrm{c}=16-13=3
\end{aligned}
$$
Put the value of $\mathrm{c}$ in equation
$$
\begin{aligned}
& 2 x+y+\mathrm{c}=0 \\
& \Rightarrow 2 x+y+3=0
\end{aligned}
$$
Now, satisfy $(1,-5)$ in the above equation.
So, $2(1)+(-5)+3=0$
$$
\begin{aligned}
& 2-5+3=0 \\
& -3+3=0
\end{aligned}
$$
Therefore, L.H.S = R.H.S
Thus, the required point is $(1,-5)$.