Finding the equation of a circle given two points and the radius 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

Let A be (6,3) and B be (7,2).
Gradient of AB = $\frac { 2-3 }{ 7-6 }$ = -1
So gradient of perpendicular bisector of AB = $-\frac { 1 }{ 1 }$ = 1
Midpoint of AB is ($\frac { 6+7 }{ 2 }$, $\frac { 3+2 }{ 2 }$), which is ($\frac { 13 }{ 2 }$, $\frac { 5 }{ 2 }$).
Hence, the equation of the perpendicular bisector is $y-\frac { 5 }{ 2 }= 1(x-\frac { 13 }{ 2 })$
The equation can be written as $y=x-4$.
Let the centre of the circles be (x,y).
Substitute $y=x-4$ as the centre lies on the perpendicular bisector.
Then the centre becomes $(x,x-4)$.
Take the distance between any point on the circle and the centre, which is the radius.
${ (x-6) }^{ 2 }+{ (x-4-3) }^{ 2 }={ \sqrt { 5 }  }^{ 2 }\\ { (x-6) }^{ 2 }+{ (x-7) }^{ 2 }=5\\ { x }^{ 2 }-12x+36+{ x }^{ 2 }-14x+49=5\\ 2{ x }^{ 2 }-26x+80=0\\ (x-8)(x-5)=0\\ So\quad x=5\quad or\quad x=8\\ As\quad y=x-4,\\ y=1\quad or\quad y=4\\ \\ $
Hence, the possible coordinates of centres are (5,1) and (8,4).
The equations are
${ (x-5) }^{ 2 }+{ (y-1) }^{ 2 }=5$
and ${ (x-8) }^{ 2 }+{ (y-4) }^{ 2 }=5\\$ 
