Intersection of an elliptic paraboloid and a sphere Problem: Prove that sphere $x^2+y^2+z^2=50z$ and elliptic paraboloid $\frac{x^2}{25}+\frac{y^2}{16}=2z$ intersects in two circles and find the radius and center of these circles.
I figured out that intersection of them lies in two planes $3y\pm4z$. Since every plane section of a sphere is a circle, we got what we wanted to prove (am I right?)
After that I decided to express $y$ in terms of $z$ and substitute y into the equation of the sphere to try find center and radius of circles. After various transformations, I got this equation: $z^2-18z-\frac{9x^2}{25}=0$. And here I stuck and don't understand how to get to the radius and center.
 A: $x^2 + y^2 + z^2 = 50 z $ and $ \dfrac{x^2}{25} + \dfrac{y^2}{16} = 2 z $
Multiplying the second equation by $25$ and subtracting, we get
$ y^2 \bigg(1 - \dfrac{25}{16} \bigg) + z^2 = 0 $
Multiply through by $16$:
$ - 9 y^2 + 16 z^2 = 0 $
which is, as you said in the question statment, the equation of the two planes
$ (4 z - 3 y) (4 z + 3 y) = 0 $
So the two planes are $4z - 3y = 0 $ and $ 4z + 3 y = 0 $
The intersection of a plane as stated is a circle, whose center is the foot of the perpendicular from the center of the sphere to the plane.
From $x^2 + y^2 + z^2 = 50 z $ we get $ x^2 + y^2 + (z - 25)^2 = 625 = 25^2 $
Hence the center of the sphere is $(0, 0, 25) $.  The normal to the first plane is $(0, -3, 4) $.  Therefore, the foot of the perpendicular to the plane from the center of the sphere is
$ C = (0, 0, 25) + t (0, -3, 4) $
$C$ lies in the plane, so  $(0, -3, 4) \cdot (0, -3 t, 25 + 4 t) = 0 $
From which $ t (9 + 16) = -100 $, therefore, $t = -4 $ which implies that
$ C = (0, 0, 25) - 4 (0, -3, 4) = (0, 12, 9 ) $
The distance between $C$ and the center of the sphere is
$ d = | t | \| (0, -3, 4 ) \| = 4 (5) = 20 $
Hence the radius of the circle is $\sqrt{ 25^2 - 20^2} = 15$
For the second plane $  4 z + 3 y = 0 $, the normal to the plane is $(0, 3, 4) $
$ C = (0, 0, 25) + t (0, 3, 4) $
$C$ lies in the plane, so  $(0, 3, 4) \cdot (0, 3 t, 25 + 4 t) = 0 $
From which $ t (9 + 16) = -100 $, therefore, $t = -4 $ which implies that
$ C = (0, 0, 25) - 4 (0, 3, 4) = (0, -12, 9 ) $
The distance between $C$ and the center of the sphere is
$ d = | t | \| (0, 3, 4 ) \| = 4 (5) = 20 $
Hence the radius of the second circle is $\sqrt{ 25^2 - 20^2} = 15$
