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 the answer of Van Snyder. However, I think it can be simplified.

Given the points $A$ and $B$ and the radius $r$ of the circle, we can determine the mid-point $p\_0$ between $A$ and $B$ with $p\_0 = \frac{1}{2}(A+B)$. Now the vector $\mathbf{v}$ from $A$ to mid-point $p\_0$ is $\mathbf{v} = p\_0 - A = \frac{1}{2}(B-A)$.

The normal $\mathbf{n}$ could be figured out by simply using the [perp operator](https://www.xarg.org/book/linear-algebra/2d-perp-operator/), without the double cross-product: $\mathbf{n}=\mathbf{v}^\perp$ and to make further processing simpler, we normalize $\mathbf{n}$, denoted as $\mathbf{\hat{n}}$.

The line $p(t)$ hitting the center $M$ of the circle is

$$p(t) = p\_0 + t \mathbf{\hat{n}}= p\_0 + t \mathbf{\hat{v}}^\perp$$

But now from pythagorean theorem we know $r^2 = |\mathbf{v}|^2 + t^2$, which also finds

$$t = \pm\sqrt{r^2-|\mathbf{v}|^2}$$

Bringing everything together, the mid-points $M$ are then

$$\begin{array}{rl}
M &= \frac{1}{2}(A+B)\pm\sqrt{r^2-|\mathbf{v}|^2}\cdot \mathbf{\hat{v}}^\perp\\
&= \frac{1}{2}(A+B)\pm\sqrt{\frac{r^2}{|\mathbf{v}|^2}-1}\cdot \mathbf{{v}}^\perp\\
\end{array}$$