Find all circles given two points and not the center This is probably pretty straight forward for you.
I have two points on a circle, $(-4, 7)$ and $(-5, 0)$. Given these two points and the radius $5$, what are all the possible equations?
My first idea was to solve the system
$\sqrt{(-4-a)^2 + (7-b)^2} = 5\tag 1$
$\sqrt{(-5-a)^2 + b^2} = 5\tag 2$ 
Where $(a, b)$ is the center point. However, I don't get the correct answers. Maybe this question should be about how to solve the system.
Thanks in advance!
Edit:
This is the formulation of the problem:
Find all circles through the points $(-4, 7)$ and $(-5, 0)$ with a radius of $5$.
 A: Following your idea:
$$(a+4)^2+(b-7)^2=5^2=(a+5)^2+b^2\implies 8a+16-14b+49=10a+25\implies$$
$$2a=-14b+40\implies a=-7b+20$$
So all the possible centers of circles fulfilling the condition are on the above straight line...
A: I'd personally start with the following 2 equations.
$(-4-a)^2 + (7-b)^2 = 25$ (1)
$(-5-a)^2 + b^2 = 25$ (2)
distribute out the squares.
$16 + 8a +a^2 + 49 - 14b + b^2 = 25$ (1)
$a^2 + b^2 + 8a - 14b + 40 = 0$ (1)
$25 + 10a + a^2 + b^2 = 25$ (2)
$a^2 + b^2 + 10a = 0$ now we set them equal
$a^2 + b^2 + 10a = a^2 + b^2 + 8a - 14b + 40$
$2a = -14b + 40$
$a = 20 -7b$ now we plug this into an earlier equation. I'm using equation 2
$(20 - 7b)^2 + b^2 + 10(20 - 7b) = 0$
$400 - 280b + 49b^2 + b^2 + 200 - 70b = 0$
$600 -350b + 50b^2 = 0$
Using the quadratic formula, b equals 4 or 3, so a = -8 or -1 respectively, giving us two possible centers of (-8,4) or (-1,3). This can be verified using the original equation.
A: Here's a more geometric approach.  Let $P=(-4, 7)$ and $Q = (-5, 0)$ be the two points on the circle.  Then, the center of the circle is a point $S$ such that $SP = SQ$.  The locus of all such points will be the perpendicular bisector of $PQ$.  This can be seen using similar triangles: Let $R$ be the midpoint of $PQ$, and $X$ be some point on the perpendicular bisector of $PQ$ other than $R$ (if $X=R$, the $PX = QX$ by definition). then $P\hat{R}X = Q\hat{R}X$, $PR=QR$, and the side $RX$ is shared between the triangles $PRX$ and $QRX$, so the triangles are similar.  Thus, $PX=QX$, and the two points lie on the same circle with center $X$.  
Using this approach lets you find $b$ in terms of $a$, so you only need to solve the a single quadratic equation (personally, I'd choose the one involving the point $(-5, 0)$).
A: Another geometric approach, letting $P=(-4,7)$ and $Q=(-5,0)$  is to construct the straight line $L$ through $P$ perpendicular to $PQ$, noting that $PQ$ has length $\sqrt {50}$. 
If $R$ is a point on $L$ distance $10$ from $Q$ then $QR$ will be a possible diameter for the circle (the angle at $P$ will be a right angle).
It just so happens that the triangle $PQR$ is isosceles right-angled (simple application of Pythagoras), so it is simple to locate $R$ (draw a diagram if it is at all difficult) - you need to add/subtract 7 from the first co-ordinate of $P$ and subtract/add 1 to the second co-ordinate giving $R=(3,6)$ or $(-11,8)$. [the vector $PQ$ is $(-1,-7)$ and we need a vector the same length at right angles]
The centre of the circle is the midpoint of $QR$ which is simply calculated as $(-1,3)$ or $(-8,4)$ respectively.
Note that this was done with very simple arithmetic and without solving any quadratic equations. However some of the other methods suggested will work better in general situations. This method takes advantage of special features of this situation.
A: Given two points ( A ( x1 , y1 ) and B ( x2 , y2 ) and radius(r).
we can find the center as follows:
M(a,b) = ( ( x1 + y1 ) / 2 , ( y1 + y2 ) / 2 ) be the midpoint of A and B.
Let distance between A and M be h.
From this midpoint, the distance to centre of required circle is 
d = ( ( r^2 ) - ( ( h^2 ) ).
The centre would be:
C ( x , y ) = ( a + ( d * ( y1 - y2 ) / ( h ) ) , b + ( d * ( x1 - x2 ) / h )
( OR ) 
C ( x , y ) = ( a - ( d * ( y1 - y2 ) / ( h ) ) , b - ( d * ( x1 - x2 ) / h )
because there can be two centres with given Radius, either side of the line joining given two points..
