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: 
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 ${ (x-8) }^{ 2 }+{ (y-4) }^{ 2 }=5\\$
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.