Finding the equation of a circle given two points and the radius 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

Here is a mostly 'square free' approach:
The two constraints are $\|x-(6,3)\|^2 = 5$ and $\|x-(7,2)\|^2 = 5$. Expanding gives $\|x\|^2-2 \langle (6,3), x \rangle +45 = 5$ and $\|x\|^2-2 \langle (7,2), x \rangle +53 = 5$,
subtracting and simplifying gives $\langle (1,-1), x \rangle = 4$. Note that this is a line at $45 °$.
The bisector of the two points is $(\frac{13}{2}, \frac{5}{2})$. The distance between the two points is $\sqrt{2}$, hence the distance from the points to the bisector is $\frac{1}{\sqrt{2}}$. Pythagoras gives the distance from the bisector to the centre of the circles as $\sqrt{5-\frac{1}{2}} = \sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}$.
If we draw a line at $45 °$ through $(\frac{13}{2}, \frac{5}{2})$, and then mark off points that are $\frac{3}{\sqrt{2}}$ away on either side, we will find the centres. The corresponding displacements are $\pm (\frac{3}{2}, \frac{3}{2})$, hence the centres are $(\frac{13}{2}, \frac{5}{2}) \pm (\frac{3}{2}, \frac{3}{2})$, which gives the centres $(5,1), (8,4)$.
Hence the equations are
$$ (x_1-5)^2+(x_2-1)^2 = 5, \ \ (x_1-8)^2+(x_2-4)^2 = 5$$
