# Intersection between a sphere and a plane

Find all integer solutions to:

$$x+y+z=94$$

$$x^2+y^2+z^2=4506$$

So clearly we have a plane and a sphere, so their intersection forms a circle, how do I locate the points on this circle which have integer coordinates (if any exist) ?

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While you can think about this as the intersection between two algebraic sets, I hardly think this is in the spirit of the tag [algebraic-geometry]. – Asaf Karagila Feb 6 '13 at 17:16

The intersection of the equations $$x + y + z = 94$$ $$x^2 + y^2 + z^2 = 4506$$ is indeed the intersection of a plane and a sphere, whose intersection, in 3-D, is indeed a circle, but if we project the circle onto the x-y plane, we can view the intersection not, per se, as a circle, but rather an ellipse:
When graphed as an implicit function of $x, y$ given by $$x^2+y^2+(94-x-y)^2=4506$$ gives us:
Hint: there are only 6 integer solution pairs $(x, y)$ that are solutions to the equation of the ellipse (the intersection of your two equations): all of which are such that $x \neq y$, $x, y \in \{1, 37, 56\}$. You can find the corresponding value of $z$ for each integer pair $(x,y)$ by solving for $z$ using the given $x, y$ and the equation $x + y + z = 94$.
Nitpick: the intersection is a circle, but its projection on the $xy$-plane is an ellipse. – mrf Feb 6 '13 at 17:17
it means, essentially, $(1, 37), (1, 56), (37, 1), (37, 56), (56, 1), (56, 37)$ are all integer solutions $(x, y)$ to the intersection. You need only find the corresponding $z$ coordinate, using the given values for $(x, y)$, using the equation $x + y + z = 94$ – amWhy Feb 6 '13 at 18:17