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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 $2x+y+6=0$?

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    $\begingroup$ The line $x - 2y + 8 = 0$ has slope $\frac{1}{2}$, while the line $2x + y + 6 = 0$ has slope $-2$. Did you mean to say the lines are perpendicular? $\endgroup$ – N. F. Taussig Jan 14 '16 at 13:51
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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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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 :enter image description here

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.

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Hint: The lines are perpendicular, and the center of the circle must be in the bisector of the angle they form.

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