Question

Sketch the \(x y\) -trace of the sphere. $$ x^{2}+y^{2}+z^{2}-6 x-10 y+6 z+30=0 $$

Short answer

The xy-trace of the sphere is a circle with center at (3, 5) and radius 4.

Step-by-step solution

Set z equal to 0

First, we need to set \(z = 0\) in the given equation. This produces the following equation: \(x^{2}+y^{2}-6 x-10 y+30=0\)

Reformulate the equation

Try to rewrite our equation and group the x terms and y terms together: \((x^{2}-6x)+(y^{2}-10y)=-30\)

Completing the square

For both x and y terms we complete the square by adding \((\frac{-6}{2})^{2}=9\) for the x-terms and \((\frac{-10}{2})^{2}=25\) for the y-terms, both to the left side and the right side of the equation. We get then \((x-3)^{2}+(y-5)^{2}=16\)

Recognize the form

We can see now that this equation is of a circle in standard form \((x - h)^{2}+(y - k)^{2}=r^{2}\), where h = 3 and k = 5 are the center of the circle and r = 4 is the radius