Find the standard equation of the sphere with center at $( - 2,3,6 ) \text { and tangent to the }$ yz-plane.
A) $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 4$
B) $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 9$
C) $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 45$
D) $( x - 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z + 6 ) ^ { 2 } = 4$
E) $( x - 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z + 6 ) ^ { 2 } = 45$

Question

Find the standard equation of the sphere with center at $( - 2,3,6 ) \text { and tangent to the }$ yz-plane. 
A) $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 4$
B) $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 9$
C) $( x + 2 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 45$
D) $( x - 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z + 6 ) ^ { 2 } = 4$
E) $( x - 2 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z + 6 ) ^ { 2 } = 45$

Quizplus · Accepted Answer

The sphere is tangent to the yz-plane at x = 0, so the distance from the center to the yz-plane is the radius. The center is at (-2,3,6), so the radius is |x-coordinate of the center| = |-2| = 2. The standard equation of a sphere is $(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$, where $(h,k,l)$ is the center and $r$ is the radius. Substituting $h = -2$, $k = 3$, $l = 6$, and $r = 2$, we get $(x + 2)^2 + (y - 3)^2 + (z - 6)^2 = 4$.

Find the Standard Equation of the Sphere with Center At (−2,3,6) and tangent to the ( - 2,3,6 ) \text { and tangent to the }(−2,3,6) and tangent to the

Find the Standard Equation of the Sphere with Center At $( - 2,3,6 ) \text { and tangent to the }$