Equation of a sphere

Question

How to find the equation of a sphere which has got the intersection of another given sphere with a given plane as its great circle? I am not able to find the equation of circle of intersection of the given sphere and plane.

Answer 1

Let us take an example : consider the sphere $S$ with center $O=(0,0,0)$ and radius $R$, therefore with equation $x^2+y^2+z^2-R^2=0$ and the plane $P$ with equation $x+y+z-1=0$.

The equation of sphere $\Sigma$ is necessarily a combination of the equations of $S$ and $P$ that can be written under the form :

$$(x^2+y^2+z^2-R^2)+k(x+y+z-1)=0 \tag{1}$$

$$ \ \iff (x+k/2)^2+(y+k/2)^2+(z+k/2)^2=R^2+k+3\frac{k^2}{4}\tag{2}$$

meaning that

$$\text{The center of} \ \Sigma \ \text{is} \ C=(-k/2,-k/2,-k/2) \ \text{and its radius} \ \rho=\sqrt{R^2+k+3\frac{k^2}{4}}.\tag{3}$$

It remains to express that the two spheres intersect along a diameter of $\Sigma$.

This is done by expressing that the center of $\Sigma$ belongs to plane $P$ (no need to find the equation of the circle), i.e.,

$$-k/2-k/2-k/2=1$$

Therefore $k=-2/3$ ; plugging this value of $k$ into (2) and (3) gives the equation of sphere $\Sigma$, its center and radius.