Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter F - Foundations: A Prelude to Functions - Section F.4 Circles - F.4 Assess Your Understanding - Page 38: 35


$x^2 + y^2 = 13$

Work Step by Step

First, we need to find the radius of the circle, so we need to see how far from the center the point given lies. Let us use the following distance formula to find the radius: $r = \sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2}$ We can now plug in our center, which is the origin $(0, 0)$, and the coordinates if the point $(-2, 3)$ into this formula: $r = \sqrt {(3 - 0)^2 + (-2 - 0)^2}$ Simplify what's in the parentheses: $r = \sqrt {(3)^2 + (-2)^2}$ Evaluate the exponents: $r = \sqrt {9 + 4}$ Simplify the radicand: $r = \sqrt {13}$ Now, we can plug in the center and the radius into the standard equation for a circle, which is given by the formula: $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the center of the circle, and $r$ is the radius. Let us plug in our values: $(x - 0)^2 + (y - 0)^2 = (\sqrt {13})^2$ Evaluate what is in parentheses first: $x^2 + y^2 = (\sqrt {13})^2$ Evaluate the exponent: $x^2 + y^2 = 13$
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