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.
The centers of two circles defined by vectors to two points $A$ and $B$, and a radius $r$, are on a line perpendicular to $B-A$ at $p\_0 = \frac12 (B+A)$, and in the plane containing $A$ and $B$.
The vector normal to the plane containing $A$ and $B$ is $n\_1 = A \times B$. The vector perpendicular to $B-A$ and in the plane is $n = n\_1 \times (B-A)$. The vector from $A$ to $p\_0$ is $v = \frac12 (B-A)$.
The centers of the circles defined by $A$, $B$, and $r$ are on the line $p(t) = p\_0 + t n = \frac12(A+B) + t n$, at a value of $t$ to be computed. The distance from either point, say $A$, to the center of the circle is $r$. Therefore
$|p(t)-A|^2 = | \frac12(B-A) + t n |^2 = | v + t n |^2 = |n|^2 t^2 + 2\, v \cdot n\, t + |v|^2 = r^2$.
The vector $n$ is (by definition) orthogonal to $v$. Therefore the second term above is zero, $|n|^2 t^2 + |v|^2 = r^2$, and $t = \pm \sqrt{ \frac{r^2-|v|^2}{|n|^2}}$.