A circle goes through $(5,1)$ and is tangent to $x-2y+6=0$ and $x-2y-4=0$. What is the circle's equation?
All I know is that the tangents are parallel, which means I can calculate the radius as half the distance between them: $\sqrt5$. So my equation is $$(x-p)^2+(y-q)^2=5$$ How can I get the locations of the centre? (I think there are 2 solutions.)
On
As the radius $=$ the perpendicular distance of a tangent from the center.
If $(h,k)$ is the center, radius $r=\dfrac{|h-2k+6|}{\sqrt{1^2+2^2}}=\dfrac{|h-2k-4|}{\sqrt{1^2+2^2}}$
Squaring we get, $$(h-2k+6)^2=(h-2k-4)^2\iff h=2k-1$$
$r=\dfrac{|-1+6|}{\sqrt{1^2+2^2}}=\sqrt5$
Finally $$(\sqrt5)^2=(h-5)^2+(k-1)^2=(2k-1-5)^2+(k-1)^2$$
$k=?$
We can place an additional constraint on the circle centre $(p,q)$. It has to lie on the line parallel to the two tangents and equidistant from them: $$p-2q+1=0\quad p=2q-1$$ Then since the circle passes through $(5,1)$: $$(5-(2q-1))^2+(1-q)^2=5$$ $$25-10(2q-1)+(2q-1)^2+1-2q+q^2=5$$ $$25-20q+10+4q^2-4q+1+1-2q+q^2=5$$ $$5q^2-26q+32=0$$ $$(q-2)(5q-16)=0$$ $$q=2\text{ or }\frac{16}5$$ Therefore the two possible centres are $(3,2)$ and $\left(\frac{27}5,\frac{16}5\right)$, leading to the circle equations $$(x-3)^2+(y-2)^2=5$$ $$\left(x-\frac{27}5\right)^2+\left(y-\frac{16}5\right)^2=5$$