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: 40

Answer

$(x - 2)^2 + (y - 2)^2 = 5$

Work Step by Step

We need to use both the midpoint formula and distance formula to find the center and the radius of this circle. Let us use the distance formula to find the radius of the circle. We are given the diameter, so if we take half of the distance of the diameter, we will have the radius. Our radius can be given by the following formula: $r = \frac{1}{2}(\sqrt {(x_2 - x_1)^2 + (y_2 - y_1)^2})$ Let us plug in our points $(4, 3)$ and $(0, 1)$ into the distance formula: $r = \frac{1}{2}(\sqrt {(4 - 0)^2 + (3 - 1)^2})$ Simplify what is in the parentheses: $r = \frac{1}{2}(\sqrt {(4)^2 + (2)^2})$ Evaluate the exponents: $r = \frac{1}{2}(\sqrt {16 + 4})$ Simplify the radicand: $r = \frac{1}{2}(\sqrt {20})$ Rewrite the radicand as the product of a perfect square and another factor: $r = \frac{1}{2}(\sqrt {4 \cdot 5})$ Take the perfect square out of the radicand: $r = \frac{1}{2}(2\sqrt {5})$ Simplify: $r = \sqrt {5}$ Now let us use the midpoint formula to find the center of the circle. The midpoint formula is given as: $M = (\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$ Let us plug in the endpoints of the diameter of our circle: $M = (\frac{4 + 0}{2}, \frac{3 + 1}{2})$ Simplify the numerators: $M = (\frac{4}{2}, \frac{4}{2})$ Simplify the fractions: $M = (2, 2)$ 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: $(x - 2)^2 + (y - 2)^2 = (\sqrt {5})^2$ Evaluate the exponent: $(x - 2)^2 + (y - 2)^2 = 5$
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