Precalculus (6th Edition)

Published by Pearson
ISBN 10: 013421742X
ISBN 13: 978-0-13421-742-0

Chapter 2 - Graphs and Functions - 2.2 Circles - 2.2 Exercises: 25

Answer

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

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 $(-2, 0)$ and $(-2, 4)$. Note that the midpoint of these two points is $(-2, 2)$. Thus, the circle has its center at $(-2, 2)$. The radius of the circle can be found by finding the distance from the center $(-2, 0)$ to $(-2, 2)$, which is $2$ units. Thus, the radius of the circle is $2$. (a) The circle has its center at $(-2, 2)$ and has a radius of $2$ units. Therefore, the equation of the circle in center-radius form is: $(x-(-2))^2+(y-2)^2=2^2 \\\color{blue}{(x+2)^2+(y-2)^2=4}$ (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: $(x+2)^2+(y-2)^2=4 \\x^2+4x+4+y^2-4y+4=4 \\x^2+y^2+4x-4y+8=4 \\x^2+y^2+4x-4y+8-4=0 \\\color{red}{x^2+y^2+4x-4y+4=0}$
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