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
$$ (x\_1-5)^2+(x\_2-1)^2 = 5, \ \ (x\_1-8)^2+(x\_2-4)^2 = 5$$
Well, the equation of a circle takes the form:

$$
(x-h)^2 + (y-k)^2 = r^2
$$

where h,k are the coordinates of the center of the circle, and r is the radius. Plugging in your values for x and y, you have the two equations:

$$
(6-h)^2 + (3-k)^2 = \sqrt{5}^2
$$

and

$$
(7-h)^2 + (2-k)^2 = \sqrt{5}^2
$$

The two solutions to these equations for h,k are the centers of the two possible circles that pass through the aforementioned points.