College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding: 10

Answer

$\color{blue}{(x-1)^2+(y-2)^2=4}$

Work Step by Step

RECALL: The standard form of a circle's equation is: $(x-h)^2 +(y-k)^2=r^2$ where $r$ = radius and $(h, k)$ is the center. The given circle has its center at $(1, 2)$. Substituting $1$ to $h$ and $2$ to $k$ gives that tentative equation of the circle: $(x-1)^2+(y-2)^2=r^2$ To find the value of $r$, substitute the $x$ and $y$ coordinates of the point on the circle $(1, 0)$ to obtain: $(x-1)^2+(y-2)^2=r^2 \\(1-1)^2+(0-2)^2=r^2 \\0^2+(-2)^2=r^2 \\0+4=r^2 \\4=r^2 \\\pm\sqrt{4} = \sqrt{r^2} \\\pm\sqrt2 = r$ Since a radius cannot be negative, then $r=2$. Therefore, the standard form of the circle's equation is: $(x-1)^2+(y-2)^2=2^2 \\\color{blue}{(x-1)^2+(y-2)^2=4}$
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