### Instruction: A sphere intersects the $xy$-plane in a circle centered at $(2,4,0)$ with radius 1.  The sphere also intersects the $yz$-plane in a circle centered at $(0,4,-7),$ with radius $r.$  Find $r.$.### Response: I need to find the equation of the sphere that passes through both circles.
Since the center of the sphere must lie on the perpendicular bisectors of both circles, I can use them to find the coordinates of the center.
The perpendicular bisector of the circle in the $xy$-plane is the line $x = 2.$
The perpendicular bisector of the circle in the $yz$-plane is the line $y = 4.$
Therefore, the center of the sphere must have the form $(2,4,z).$
To find $z,$ I can use the fact that the distance from the center to any point on the sphere is constant.
Let $D$ be the radius of the sphere.
Then, by the Pythagorean theorem, I have $D^2 = 1^2 + (z-0)^2$ from the circle in the $xy$-plane, and $D^2 = r^2 + (z+7)^2$ from the circle in the $yz$-plane.
Equating these two expressions, I get $1^2 + (z-0)^2 = r^2 + (z+7)^2.$
Expanding and simplifying, I get $z^2 + 1 = r^2 + z^2 + 14z + 49.$
Subtracting $z^2$ from both sides, I get $1 = r^2 + 14z + 49.$
Rearranging, I get $r^2 = -14z - 48.$
Since $r^2$ must be positive, I need to find a value of $z$ that makes $-14z - 48$ positive.
One such value is $z = -4.$
Then, $r^2 = -14(-4) - 48 = 8.$
Taking the positive square root, I get $r = \sqrt{8} = 2\sqrt{2}.$