## Trigonometry (11th Edition) Clone

Published by Pearson

# Appendix B - Graphs of Equations - Exercises - Page 427: 47

#### Answer

Refer to Graph I, Center of the circle = $(3, 1)$ Radius = $\frac{5 – 1}{2}$ = $2$ The center-radius form of the circle equation is $(x – 3)^2$ + $(y – 1)^2$ = $4$

#### Work Step by Step

Refer to Graph I, Center of the circle = $(3, 1)$ Radius = $\frac{5 – 1}{2}$ = $2$ The center-radius form of the circle equation is $(x – 3)^2$ + $(y – 1)^2$ = $2^2$ $(x – 3)^2$ + $(y – 1)^2$ = $4$

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