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 $2x+y+6=0$?
 A: 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 $h$ & $k$ 

A: let  $(d)$ be the line of equation $ x-2y+8=0$.
let  $(d')$ be the line of equation $ 2x+y+6=0$.
Note that slope of  $(d)$ is  $\frac{1}{2}$ and  slope of  $(d')$ is  $-2$, then  $(d)$  and $(d')$  are perpendicular. Then clearly the center of the cercle belongs to the line  $D$ of the bisector of the angle between  $d$  and  $d'$. 
To find the equation of this st line $D$, consider the intersecting point of  $d$ and  $d'$, this point is  $( -4,2) $ belongs  to  $D$.  
Now to find the slope of  $D$ enough to recognize that  the angle between line $D$  and the x-axis  is the sum of the angles between  $D$ and  $d$ and the angle between  $d$ and  $x-axis$(let us call this final point by  $\theta$). then the slope of  line  $D$ is  $$ \tan(45+\theta)= \frac{\tan(45)+\tan(\theta)}{1-\tan(45)\tan(\theta)}$$
But  $\tan(\theta)=\frac{1}{2}$ the slope of the line  $d$, so the slope of  the line  $D$  is  $$ \frac{1+\frac{1}{2}}{1-\frac{1}{2}}=3$$ So the equation of line $D$ is  $y=3x+14$
Now consider the following plot of the lines :
If we suppose  the unkown point $C$ is the center of your circle, then as this circle is tangent to the line $d$ , then the angle at  $A$ is right. So consider the right triangle  $CAI$,  we hvae  $$\sin(45)=\frac{AC}{IC}=\frac{1}{\sqrt{2}}$$  but  $AC$ is equal to $GC$ since both are radius of the circle. So we get  $$ IC=\sqrt{2}GC$$ Recall that   $I=(-4,2)$ and  $G=(-1,6)$ (given)  and  $C $ belongs to the line $D$ so it satisfy  $y=3x+14$, so substitute all these in the last above equation,  you will get one equation in the unknown  $x$ . Solve for $x$ then  find  $y=3x+14$ and this is the center of your circle.
Hope this helps.
A: Hint: The lines are perpendicular, and the center of the circle must be in the bisector of the angle they form.
A: We know that if two tangents are constructed on the two ends of a diameter, they are always parallel. Hence, by conversing this, we get a diameter if we draw two parallel tangents in a circle. To find the center, we can make two diameters by the same method, one horizontal and the other vertical.
As show in the figure

