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Chapter 13: Problem 53

Sketch the \(y z\) -trace of the sphere. $$ x^{2}+y^{2}+z^{2}-4 x-4 y-6 z-12=0 $$

Short Answer

Expert verified
The \(yz\)-trace of the given sphere is a circle on the \(yz\)-plane with the equation \((y - 2)^{2} + (z - 3)^{2} = 12\), centered at point \((2,3)\) with a radius of \(\sqrt{12}\).

Step by step solution

01

Convert to standard form

Get the equation of the sphere into its standard form. Complete the square on the \(x\), \(y\), and \(z\) terms. Begin by rewriting and regrouping the terms: \(x^{2}-4 x + y^{2} - 4y + z^{2} - 6z = 12\). Now complete the squares: \((x - 2)^{2} + (y - 2)^{2} + (z - 3)^{2} = 16\). Thus, the center of the sphere is at \((2, 2, 3)\) and its radius is \(\sqrt{16} = 4\).
02

Find the Circle equation

Find the equation of the \(yz\)-trace by setting \(x = 0\). This yields \((0 - 2)^{2} + (y - 2)^{2} + (z - 3)^{2} = 16\), which simplifies to \(4 + (y - 2)^{2} + (z - 3)^{2} = 16\). Thus, the equation of the \(yz\)-trace is \((y - 2)^{2} + (z - 3)^{2} = 12\). With center at \((2,3)\) and radius is \(\sqrt{12}\).
03

Sketch the Circle

The final step is to sketch the circle on the \(yz\)-plane. The center is at point \((2,3)\) and the radius of the circle is \(\sqrt{12}\). Draw a circle centered at \((2,3)\) with a radius of \(\sqrt{12}\) on the \(yz\)-plane.

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