## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - 2.2 Circles - 2.2 Exercises - Page 200: 21

#### Answer

$\color{blue}{(x-\sqrt2)^2+(y-\sqrt2)^2=2}$ Refer to the graph below.

#### Work Step by Step

RECALL: The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units. Thus, the given circle's equation in center-radius form is: $(x-\sqrt2)^2+(y-\sqrt2)^2=(\sqrt2)^2 \\\color{blue}{(x-\sqrt2)^2+(y-\sqrt2)^2=2}$ To graph the circle, perform the following steps: (1) Plot the center (note that $\sqrt2 \approx 1.414$). (2) Plot the points $\sqrt2$ units above, below, to the left, and to the right of the center. (3) Connect the four points in Step 2 using a curve to form a circle. (Refer to the attached image in the answer part above for the graph.)

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