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: 
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
$$ (x\_1-5)^2+(x\_2-1)^2 = 5, \ \ (x\_1-8)^2+(x\_2-4)^2 = 5$$
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 $(5,1)$