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.
I like your approach

The slope of the line between the two points is $\frac {7-6}{2-3} = -1$  

The distance between the points is $\sqrt {(7-6)^2 + (2-3)^2} = \sqrt 2$  

The midpoint is $(\frac {13}{2}, \frac {5}{2})$  

The distance from either point to the midpoint is $\frac {\sqrt 2}{2}$

We need to find points on the line of the perpendicular bisector that are the correct distance from the given points.

Using the Pythagorean theorem:  

$(\frac {\sqrt 2}{2})^2 + D^2 = (\sqrt 5)^2\\D^2 = \frac {9}{2}\\D = \frac {3\sqrt 2}{2}$

Each time we step $(\frac 12, \frac 12)$ from the midpoint we are getting $\frac {\sqrt 2}{2}$ closer to the centers that we are looking for.

$(\frac {13}{2}, \frac {5}{2}) + (\frac 32,\frac 32)$ or $(\frac {13}{2}, \frac {5}{2}) - (\frac 32,\frac 32)$

$(8,4)$ or $(5,1)$