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Chapter 13: Problem 52

Sketch the \(y z\) -trace of the sphere. $$ (x+2)^{2}+(y-3)^{2}+z^{2}=9 $$

Short Answer

Expert verified
The \(y z\)-trace of the sphere \( (x+2)^{2}+(y-3)^{2}+z^{2}=9 \) is a circle in the \(y z\) -plane with center at \((y, z) = (3, 0)\) and radius \(\sqrt{5}\).

Step by step solution

01

Set x to 0

Substitute \(x = 0\) into the equation of the sphere. The equation becomes \((0 + 2)^{2} + (y - 3)^{2} + z^{2} = 9\). This simplifies down to \(4 + (y - 3)^{2} + z^{2} = 9\). We then isolate for \(z^{2}\), the equation further becomes \((y - 3)^{2} + z^{2} = 9 - 4 = 5\).
02

Recognize and Sketch the Geometric Form

The equation \((y - 3)^{2} + z^{2} = 5 \) represents a circle in the \(y z\) -plane with center at (3, 0) and radius \( \sqrt{5} \). This is recognized as the general form of the equation for a circle in two dimensions, \((y - h)^{2} + (z - k)^{2} = r^{2}\), where the center of the circle is \((h, k)\) and \(r\) is the radius.
03

Sketch the \(y z\)-trace

Sketch the circle on the \(y z\) -plane with the recognized center and radius. The center point is at \((y, z) = (3, 0)\) and the radius is \(\sqrt{5}\).

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