Finding the position vector of the centre of a circle of intersection. If there are two spheres with vector equations:
$(r-a)^2=A^2$ and $(r-b)^2=B^2$
where, intuitively, "a" and "b" are vectors representing the centre of the spheres and "A" and "B" are scalars representing the radius of the spheres,
Assuming that they intersect each others, how to show that the position vector of the centre of the circle of intersection is given by:
$$\frac{a+b}{2} + \frac{(A^2-B^2)(b-a)}{2(b-a)^2}$$
 A: Consider any plane containing the sphere's centers $\mathbf a$ and $\mathbf b$. If there is an intersecting circle, it will also contain two points from this circle. Let us consider one and call it $\mathbf c$.
This triangle is illustrated here as follows:

As you can see, the sides of the triangles coincide with the radii of the spheres ($A$ and $B$) and the connection of the sphere's center ($\left\|\mathbf b - \mathbf a\right\|$). Moreover, the altitude of the triangle coincides with a radius of the interseting circle (denoted by $R$) and the altitude's foot is the center of the circle. I denote it by $\mathbf x$. Since $\mathbf x$ is on the line through $\mathbf a$ and $\mathbf b$, we can express it as $\mathbf x = \mathbf a + t\cdot(\mathbf b - \mathbf a)$, where $t \in (0,1)$. We therefore have $\left\|\mathbf x - \mathbf a\right\| = t \cdot \left\|\mathbf b - \mathbf a\right\|$ and  $\left\|\mathbf b - \mathbf x\right\| = (1-t) \cdot \left\|\mathbf b - \mathbf a\right\|$
To compute $t$ consider the two right triangles that have the altitude in common. From Pythagoras we have $$R^2 = A^2 - t^2\left\|\mathbf b - \mathbf a\right\|^2 = B^2-(1-t)^2\left\|\mathbf b - \mathbf a\right\|^2.$$
Solving for $t$ gives $$A^2-B^2 = (t^2-1+2t-t^2)\left\|\mathbf b - \mathbf a\right\|^2 \quad \Rightarrow t = \frac{A^2-B^2}{2\left\|\mathbf b - \mathbf a\right\|^2} + \frac{1}{2}.$$
Finally, since  $\mathbf x = \mathbf a + t\cdot(\mathbf b - \mathbf a)$, the result is $$\mathbf x = \mathbf a + \frac{A^2-B^2}{2\left\|\mathbf b - \mathbf a\right\|^2} (\mathbf b - \mathbf a) + \frac{1}{2}(\mathbf b - \mathbf a) = \frac{\mathbf a + \mathbf b}{2} + \frac{A^2-B^2}{2\left\|\mathbf b - \mathbf a\right\|^2} (\mathbf b - \mathbf a).$$
