Given equation, $y^2-6y-4x+13=0$

Now origin is shifted to (h,k)

$y\to y+k 2x\to x+h$

$k(y+k{)}^2-6(y+k)-4(x+h)+13=0$

$y^2+k^2+2ky-6y-6k-4x-4h+13=0$

$y^2$ will not contain any term

$y^2+2k-6y=0 \& k^2-6k-4h+13=0$

$$\begin{gathered} y(y+2k-6)=0 (3{)}^2-6(3)-4h+13=0 \\ y=0 /y+2k-6=0 9-18-4h+13=0 \\ 2k=6 -4h+4=0 \\ k=3 \cancel{ -4}h=\cancel{-4} \\ h=1 \end{gathered}$$

Origin shifted to $(h,k)\to (1,3)$.