find equation of the middle circle The diagram below is of 3 circles have 3 centres A, B and C and they are collinear.

The equations of the circumferences of the outer circles are 
${(x + 12)^2 + (y + 15)^2 = 25}$
and
${(x - 24)^2 + (y - 12)^2 = 100}$
The question is to find the equation of the centre circle.
I can tell the centres of the 2 circles are (-12, -15) and (24, 12)
Using the distance formula I can tell that the distance between AC is 45 and therefore the radius is 15 by subtracting the other 2 radii.
I am not sure how to get the coordinates of the centre circle.  It is not the midpoint.
The gradient is ${3\over 4}$.  Should I use this to somehow use the ratio to find the centre of the centre circle?
 A: $\bar{AB} = 5 + 15 = 20$ and so $B$ is $20/45 = 4/9$ of the way up, so it must be that its coordinates are $B_x = -12 + 36*4/9$ and $B_y = -15 + 27*4/9$.
A: Hint: We know from the equations that
$$
A=(-12,-15)\qquad\text{and}\qquad C=(24,12)
$$The distance between $A$ and $C$ is $\sqrt{36^2+27^2}=45$.
The radius of the circle around $A$ is $5$.
The radius of the circle around $C$ is $10$.
That leaves $30$ as the diameter of the circle around $B$.

Therefore, the distance from $A$ to $B$ is $5+15=20$ and the distance from $B$ to $C$ is $10+15=25$.
Thus,
$$
\left(C_x-B_x\right)+\left(B_x-A_x\right)=C_x-A_x=36
$$
Furthermore, the distance in $x$ between the centers is in the same ratio as the distance between the centers:
$$
\frac{C_x-B_x}{B_x-A_x}=\frac{25}{20}
$$
The same is true for the $y$ coordinates:
$$
\left(C_y-B_y\right)+\left(B_y-A_y\right)=C_y-A_y=27
$$
and
$$
\frac{C_y-B_y}{B_y-A_y}=\frac{25}{20}
$$
Now you can compute $B_x$ and $B_y$.
A: Here is a simple geometric method to find the coordinates of $B$ using similar right triangles (some of the other answers use this but do not explain that they are doing so and skip some steps).
So we know that point $A$ is at $(-12, -15)$ and point $C$ is at $(24,12)$. Also observe that the radius of circle $A$ is $5$ and the radius of circle $C$ is $10$. As you said, the distance is $45$, and so the diameter of circle $B$ must be $30$ (yielding the more important radius of $15$). We then imagine a right triangle with hypotenuse $\overline{AB}$ and a horizontal leg parallel with the $x$-axis (the other leg is obviously parallel with the $y$-axis). We then notice that the two triangles are similar. We thus conclude that $${\;\;\overline{AB}\;\; \above 1pt \;\;\overline{AC}\;\;} = {\;\;\overline{AB}_x\;\; \above 1pt \;\;\overline{AC}_x\;\;} \qquad \text{and} \qquad {\;\;\overline{AB}\;\; \above 1pt \;\;\overline{AC}\;\;} = {\;\;\overline{AB}_y\;\; \above 1pt \;\;\overline{AC}_y\;\;}$$
We can see that $\overline{AB} = 20$, $\overline{AC} = 45$, and $\overline{AC}_x = 36$ from our calculations above, and a little arithmetic yields $\color{red}{\overline{AB}_x = 16}$. Using $\overline{AC}_y = 27$, we then find that $\color{red}{\overline{AB}_y = 12}$.   
We get $B_x$ by taking $A_x + \overline{AB}_x = -12 + 6 = \color{red}{4 = B_x}$
We get $B_y$ by taking $A_y + \overline{AB}_y = -15 + 12 = \color{red}{-3 = B_y}$.
Thus, we get our answer by using the coordinates of the center and the radius, yielding
$$\color{red}{(y+3)^2 + (x-4)^2 = 15^2}$$
Here's a picture I threw together in paint to clarify (hopefully... not the greatest drawing!)

