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!
Best Answer
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.