Question

find the center and radius of the circle which is the intersection of the sphere x^2+y^2+z^2 -4x+2y-6z-11=0 which are parallel to the plane x=0​

Answer 1

An equation of the sphere with center (3, −11, 6) and radius 10 is (x − 3)

2 +

[y − (−11)]2 + (z − 6)

2 = 102 or (x − 3)

2 + (y + 11)

2 + (z − 6)

2 = 100. The

intersection of this sphere with the xy-plane is the set of points on the sphere

whose z-coordinate is 0. Putting z = 0 into the equation, we have (x − 3)

2 +

(y + 11)

2 = 64, z = 0 which represents a circle in the xy-plane with center

(3, −11, 0) and radius 8. To find the intersection with the xz-plane, we set

y = 0: (x − 3)

2 + (z − 6)

2 = −21. Since no points satisfy this equation,

the sphere does not intersect the xz-plane. (Also note that the distance

from the center of the sphere to the xz-plane is greater than the radius

of the sphere.) To find the intersection with the yz-plane, we set x = 0:

(y + 11)

2+(z − 6)

2 = 91, x = 0, a circle in the yz-plane with center (0, −11, 6)

and radius √

91.