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Prove that the intersection of a sphere in a plane is a circle.


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.

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In fact, I think you basically have it. By construction, these are right triangles, you know the points are a radius $r$ from the origin, and you know the two triangles share a leg (and hence have the same length). Now apply Pythagorean theorem so you have two triangles that are congruent, thus every point on the points of intersection must be the same distance away from $X$. Since you're in a plane, this is a circle. –  Clayton Feb 13 '13 at 5:22
Please see this problem, where the intersection of a plane and a sphere is an ellipse: math.stackexchange.com/questions/296385/… –  Ron Gordon Feb 13 '13 at 10:08
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1 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 $r = \sqrt{R^2 - z_0^2}$

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