College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.2 - Circles - 2.2 Exercises - Page 186: 34

Answer

$3\sqrt{5}$ units

Work Step by Step

To find the distance between the two points, we use the distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ Here we use the two points $(-1,3)$ and $(5,\ -9)$: $d=\sqrt{(5--1))^{2}+(-9-3)^{2}}$ $=\sqrt{6^{2}+(-12)^{2}}$ $=\sqrt{36+144}$ $=\sqrt{180}$ $=\sqrt{36*5}$ $=6\sqrt{5}$ This gives us the distance and hence the diameter. To find the radius, we divide in two: $radius=diameter/2$ $=\frac{1}{2}(6\sqrt{5})$ $=3\sqrt{5}$
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