## Precalculus (6th Edition)

Published by Pearson

# Chapter 2 - Graphs and Functions - 2.2 Circles - 2.2 Exercises - Page 200: 23

#### Answer

(a) center-radius form: $\color{blue}{(x-3)^2+(y-1)^2=4}$ (b) general form: $\color{red}{x^2+y^2-6x-2y+6=0}$

#### Work Step by Step

RECALL: (1) 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. (2) The general form of a circle's equation is $x^2+y^2+Dx+Ey+F=0$ The center of the circle can be found by finding the the midpoint of $(3, -1)$ and $(3, 3)$. Note that the midpoint of these two points is $(3, 1)$. Thus, the circle has its center at $(3, 1)$. The radius of the circle can be found by finding the distance from the center $(3, 1)$ to $(3, 3)$, which is 2 units. Thus, the radius of the circle is $2$. (a) The circle has its center at $(3, 1)$ and has a radius of $2$ units. Therefore, the equation of the circle in center-radius form is: $(x-3)^2+(y-1)^2=2^2 \\\color{blue}{(x-3)^2+(y-1)^2=4}$ (b) To find the equation of the circle in general form, expand the equation in (a) above then put all terms on the left side to obtain: $(x-3)^2+(y-1)^2=4 \\x^2-6x+9+y^2-2y+1=4 \\x^2+y^2-6x-2y+10=4 \\x^2+y^2-6x-2y+10-4=0 \\\color{red}{x^2+y^2-6x-2y+6=0}$

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