Question

Sketch the \(x y\) -trace of the sphere. $$ (x+1)^{2}+(y+2)^{2}+(z-2)^{2}=16 $$

Short answer

The xy-trace of the equation \((x+1)^{2}+(y+2)^{2}+(z-2)^{2}=16\) is a circle with center (-1,-2) and radius \(\sqrt{12}\) .

Step-by-step solution

Set z = 0 in the spherical equation

For the xy-trace, z=0, thus our equation becomes:\[(x+1)^{2}+(y+2)^{2}+(0 - 2)^{2}=16 \]

Simplify the equation

Simplify the equation by letting (z - 2)^2 = 4. This gives:\[(x+1)^{2}+(y+2)^{2}+4=16 \]By bringing 4 to the other side of the equation, you get:\[(x+1)^{2}+(y+2)^{2}=12\]This is the equation of a circle in the xy-plane, with center (-1,-2) and radius of \(\sqrt{12}\) , that is about two units long.

Sketch the circle in the xy-plane

Sketch a circle with center at point (-1, -2) and a radius of approximately two units. This circle represents the xy-trace of the sphere in the xy-plane.