Prove that the intersection of a sphere in a plane is a circle.
Attempt:
Let $O$ be the center of the sphere, and let $\pi$ be the plane intersecting the sphere. Construct a perpendicular line from $O$ to $\pi$ and let $X$ be the point that the perpendicular line intersects on $\pi$. Now put two points on the intersection and make two triangles. This is as far as I got. I feel like if I can prove that both triangles are congruent then it proves that every point from $O$ is the same distance which implies a circle.
Best Answer
Introduce system of coordinates $xy$ plane of which lies on a given plane that crosses a sphere. Equation of the plane is $z = 0$ then, and equation of the sphere $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2 = R^2$, so after substitution you'll find that $$ (x-x_0)^2+(y-y_0)^2 = R^2-z_0^2 $$ Also you can see that not every plane can intersect a sphere ($R < z_0$) or it might be only 1 point ($R = z_0$, then the plane is a tangent plane), and it might be a circle (if $R > z_0$) with the radius $r = \sqrt{R^2 - z_0^2}$