Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Section 1.9 - Distance and Midpoint Formulas; Circles - Exercise Set - Page 281: 78


$(x-3)^2+(y-2)^2=25$ Refer to the step-by-step part below for explanation.

Work Step by Step

The circle $(x-3)^2 + (y-2)^2=25$ is in standard form. The standard form of a circles equation is $(x-h)^2 + (y-k)^2=r^2$ where $(h,k)$ is the center and $r$ is the radius. Given an equation in standard form, the values of h and k can be determined by simply taking the "opposite" of the number you see beside x and y. To illustrate: In the given example above, since what you have are: $(x-3)^2$, it means that $h=3$, the "opposite of $-3$. $(y-2)^2$, it means that $k=2$, the "opposite" of $-2$. If the equation is $(x+3)^2 + (y-7)^2=25$, then $h = -3$ (the opposite of 3) and $k=7$ (the opposite of 7). The value of $r$ can be determined by simply taking the square root of the constant found on the right side.
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