Finding the equation of the sphere that passes through a circle and a point

Question

The problem is the following: I am given a circle by the equations $$x^2+y^2+z^2+2ax+2by+2cz+d=0, Ax+By+Cz+D=0$$ (The first equation is a sphere, the second one a plane, so as I understand the plane cuts the sphere so we get a circle). I have to find the equation of a sphere that passes through the given circle and contains the point $(x_0;y_0;z_0)$, which does not belong to the given plane.

The answer is $$(Ax_0+By_0+Cz_0+D)(x^2+y^2+z^2+2ax+2by+2cz+d)-(x_0^2+y_0^2+z_0^2+2ax_0+2by_0+2cz_0+d)(Ax+By+Cz+D)=0$$ I don't understand how this can be the answer. Why do we have to pug in that point then subtract? I tried rewriting the equation of the given sphere so that I can find the center and the radius, but nothing happened. Could you please give me some hints or methods to approach this problem? Thank you very much!

Answer 1

HINT:

The idea is: any solution of the system $$f=0\\ g=0$$ is also a solution of $f+ \lambda g =0$ for any $\lambda$. So now consider the sphere of equation $$x^2 + y^2 +z^2 + 2 a x + 2 b y + 2 c z + d + \lambda(A x + B y + C z + D)=0$$ that passes through the point $(x_0, y_0, z_0)$. $\lambda$ can be easily obtained now.

Answer 2

Consider the circumference $\mathcal C$ of equations

\begin{cases} x^2+y^2+z^2+2ax+2by+2cz+d=0 &\\ Ax+By+Cz+D=0 \end{cases}

A sphere containing $\mathcal C$ is therefore $\mathcal S_1$ of equation $x^2+y^2+z^2+2ax+2by+2cz+d=0$.

Any other sphere $\mathcal S$ containing $\mathcal C$ must have an equation of the form:

$$x^2+y^2+z^2+2a'x+2b'y+2c'z+d'=0 \tag 1$$

such that $$\scriptstyle x^2+y^2+z^2+2a'x+2b'y+2c'z+d'-(x^2+y^2+z^2+2ax+2by+2cz+d) =\lambda (Ax+By+Cz+D) \tag 2\\ $$ where $\lambda \in \Bbb R$.

If, for example, we want to determine the sphere containing $\mathcal C$ and passing through $P_0 = (x_0, y_0, z_0)$ we must choose $λ$ such that the assigned point satisfies the $$x^2+y^2+z^2+2ax+2by+2cz+d-\lambda (Ax+By+Cz+D)=0$$