Find the radius of the sphere whose equation is $x ^ { 2 } + 2 x + y ^ { 2 } + z ^ { 2 } = 4$ .
A)1
B) $\sqrt { 2 }$
C) $\sqrt { 3 }$
D)2
E) $\sqrt { 5 }$
F) $\sqrt { 6 }$
G) $\sqrt { 7 }$
H) $2 \sqrt { 2 }$

Question

Find the radius of the sphere whose equation is $x ^ { 2 } + 2 x + y ^ { 2 } + z ^ { 2 } = 4$ .
A)1
B) $\sqrt { 2 }$ 
C) $\sqrt { 3 }$ 
D)2
E) $\sqrt { 5 }$ 
F) $\sqrt { 6 }$ 
G) $\sqrt { 7 }$ 
H) $2 \sqrt { 2 }$

Quizplus · Accepted Answer

To find the radius of the sphere, we first complete the square for the $x$ term in the equation. The equation of a sphere is generally given by $(x - h)^2 + (y - k)^2 + (z - l)^2 = r^2$, where $(h, k, l)$ is the center of the sphere and $r$ is the radius. The given equation is $x^2 + 2x + y^2 + z^2 = 4$. Completing the square for the $x$ term, we add and subtract $(2/2)^2 = 1$, obtaining $x^2 + 2x + 1 + y^2 + z^2 = 4 + 1$, which simplifies to $(x + 1)^2 + y^2 + z^2 = 5$. This shows the sphere has a radius $r$ such that $r^2 = 5$, hence $r = \sqrt{5}$.

Find the Radius of the Sphere Whose Equation Is x2+2x+y2+z2=4x ^ { 2 } + 2 x + y ^ { 2 } + z ^ { 2 } = 4x2+2x+y2+z2=4

Find the Radius of the Sphere Whose Equation Is $x ^ { 2 } + 2 x + y ^ { 2 } + z ^ { 2 } = 4$