Question: Can't seem to figure this out - the question is: 

**There are exactly two circles of radius $r = \sqrt{5}$ through the points $(6,3)$ and $(7,2)$. Find the equations of both circles.** 

I was thinking that I would find the equation of the line passing through these two points which would give me a chord on the circle. I could then find a line perpendicular to this by taking the negative reciprocal. This perpendicular bisector of the chord would pass through the center of the circle (which I'm assuming I need to find). 

Am I making this more complicated than it actually is...? I don't know where to go from here

Answer: 
Alternately, note that $(h,k)$ will be the center of such a circle if and only if $$(6-h)^2+(3-k)^2=5$$ and $$(7-h)^2+(2-k)^2=5.$$ Solving this system for $h$ and $k$ (there will be two possible solutions) will give you the desired circles.
This problem is a non-linear system of two equations and two unknowns.
$$\begin{align}(6-m\_x)^2 + (3-m\_y)^2 = 5\\(7-m\_x)^2 + (2-m\_y)^2 = 5\end{align}$$
$\Leftrightarrow$
$$\begin{align}m\_x^2 - 12m\_x + m\_y^2 - 6m\_y + 40=0\tag{1}\\m\_x^2 - 14m\_x + m\_y^2 - 4m\_y + 48=0\tag{2}\end{align}$$
$(1)-(2)$ is
$$\begin{align}2m\_x - 2m\_y - 8=0\end{align}$$
or
$$\begin{align}m\_x = m\_y + 4\tag{3}\end{align}$$
inserting $(3)$ into $(1)$ gives
$$\begin{align}(m\_y+4)^2 - 12(m\_y+4) + m\_y^2 - 6m\_y + 40=\\my^2+8m\_y+16 - 18m\_y -48 + m\_y^2 + 40=\\2m\_y^2 - 10m\_y +8=0\\\end{align}$$
or
$$\begin{align}m\_y^2-5m\_y+4=0\end{align}$$
solving we get
$$\begin{align}m\_{x1}=5,m\_{y1}=1&\end{align}$$
and
$$\begin{align}m\_{x2}=8,m\_{y2}=4&\end{align}$$