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: 24

Answer

(a) center-radius form: $\color{blue}{(x+1)^2+(y+2)^2=9}$ (b) general form: $\color{red}{x^2+y^2+2x+4y-4=0}$

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 $(-1, 1)$ and $(-1, -5)$. Note that the midpoint of these two points is $(-1, -2)$. Thus, the circle has its center at $(-1, -2)$. The radius of the circle can be found by finding the distance from the center $(-1, -5)$ to $(-1, -2)$, which is 3 units. Thus, the radius of the circle is $3$. (a) The circle has its center at $(-1, -2)$ and has a radius of $3$ units. Therefore, the equation of the circle in center-radius form is: $(x-(-1))^2+(y-(-2))^2=3^2 \\\color{blue}{(x+1)^2+(y+2)^2=9}$ (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: $(x+1)^2+(y+2)^2=9 \\x^2+2x+1+y^2+4y+4=9 \\x^2+y^2+2x+4y+5=9 \\x^2+y^2+2x+4y+5-9=0 \\\color{red}{x^2+y^2+2x+4y-4=0}$
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