Question

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

Short answer

The yz trace of the sphere with equation \(x^{2}+y^{2}+z^{2}-6 x-10 y+6 z+30=0\) is a circle with center at (5,-3) and radius 3.

Step-by-step solution

Locating the parameters for the sphere

Compare the given equation \(x^{2}+y^{2}+z^{2}-6 x-10 y+6 z+30=0 \) to the standard form \(x^{2}+y^{2}+z^{2}+2gx+2fy+2hz+c=0\). Here, we see that g, f, and h will be -3, -5, and 3 respectively, and c will be 30. The center of the sphere is (g, f, h) or (3, 5, -3). The radius \(r\) is \(\sqrt{g^2+f^2+h^2-c}\), which gives \(\sqrt{(3)^2+(-5)^2+(-3)^2-30}\), resulting in a radius of 3.

Setting x component to 0

In the \(y z\) -trace, we consider x component to be zero. The equation now becomes \(0+y^{2}+z^{2}-10 y+6 z+30=0\). This simplifies to \(y^2 + z^2 - 10y + 6z +30 = 0\).

Factoring the terms with y and z

Completing the square for the terms for both y and z, we get \((y-5)^{2}+(z+3)^{2}=9\), where 5 and -3 are the y and z coordinates of the centre and 9 is the radius squared.

Sketching the yz trace

The yz trace will be a circle centred at (5,-3) and with radius 3 on the yz plane. For this, plot the center, then mark the radius length in all four directions and sketch the circle connecting these points.