College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.2 - Graphs of Equations in Two Variables; Circles - 1.2 Exercises: 77

Answer

$(x-2)^2+(y-5)^2=25$

Work Step by Step

The center of the circle is the midpoint of the diameter. Find the center (midpoint of the diameter) using the midpoint formula $\left(\dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2}\right)$ to obtain: center = $\left(\dfrac{-1+5}{2}, \dfrac{1+9}{2}\right)=\left(\dfrac{4}{2}, \dfrac{10}{2}\right)=(2, 5)$ RECALL: The standard form of a circle whose center is at $(h, k)$ and radius $r$ is $(x--h)^2+(y-k)^2=r^2$. The center is at $(2, 5)$, so $h=2$ and $k=5$. Thus, the standard form of the given circle's equation of the circle is: $(x-2)^2+(y-5)^2=r^2$ The circle passes through the point $(5, 9)$. This means that the coordinates of this point satisfy the equation of the circle. Substitute the x and y values of the point into the equation to obtain: $(x-2)^2+(y-5)^2=r^2 \\(5-2)^2+(9-5)^2=r^2 \\3^2+4^2=r^2 \\9+16=r^2 \\25=r^2$ Thus, the standard form of the given circle's equation is: $(x-2)^2+(y-5)^2=25$
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