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: 
The approach seems sensible. A quadratic equation is inevitable, and this is as good a way to get to it as any.

The slope of the chord is $-1$, so the slope of the perpendicular bisector is $1$. Thus the perpendicular bisector has equation of the shape $y=x+b$. Since the midpoint $(13/2,5/2)$ of our two given points is on the perpendicular bisector, we have $b=-4$.

Now we find the centre. Say it has coordinates $(x,x-4)$. The distance to $(6,3)$ is $\sqrt{5}$. That gives
$$(x-6)^2+(x-7)^2=5.$$
When you simplify, you will find that the quadratic even factors nicely.
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 $(5,1)$