<s>[INST] Find the center of circle given two tangent lines (the lines are parallel) and a point. How to find the center of a circle if the circle is passing through $(-1,6)$ and tangent to the lines $x-2y+8=0$ and $2x+y+6=0$?
 [/INST] Let $(h, k)$ be the center of the circle then the distance of the center $(h, k)$ & $(-1, 6)$ will be equal to the radius ($r$) of circle $$r=\sqrt{(h+1)^2+(k-6)^2}\tag 1$$
Now, the perpendicular distance of the center $(h, k)$ from the line: $x-2y+8=0$
$$r=\left|\frac{h-2k+8}{\sqrt{1^2+(-2)^2}}\right|=\left|\frac{h-2k+8}{\sqrt 5}\right|\tag 2$$
the perpendicular distance of the center $(h, k)$ from the line: $2x+y+6=0$
$$r=\left|\frac{2h+k+6}{\sqrt{2^2+1^2}}\right|=\left|\frac{2h+k+6}{\sqrt 5}\right|\tag 3$$
from (2) & (3), $$|h-2k+8|=|2h+k+6|$$
$$h-2k+8=\pm(2h+k+6)$$ $$h+3k=2\tag 4$$ or $$3h-k+14=0\tag 5$$
from (1) & (2), $$\sqrt{(h+1)^2+(k-6)^2}=\left|\frac{h-2k+8}{\sqrt 5}\right|$$ 
$$(h+1)^2+(k-6)^2=\frac{(h-2k+8)^2}{5}$$
setting $k=3h+14$ from (5), one should get 
$$(h+1)^2+(3h+14-6)^2=\frac{(h-2(3h+14)+8)^2}{5}$$
$$h^2+2h-3=0\implies h=1, -3$$
hence, corresponding values of $k$ are $k=3(1)+14=17$ & $k=3(-3)+14=5$
Hence, the center of the circle is $\color{red}{(1, 17)}$
 or  $\color{red}{(-3, 5)}$

Note: It can be checked that for $h+3k=2$ from (4), there is no real values of $h$ & $k$ 

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