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: 78

Answer

$(x-3)^2+(y+1)^2=32$

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 = $\left(\dfrac{-1+7}{2}, \dfrac{3+(-5)}{2}\right)=\left(\dfrac{6}{2}, \dfrac{-2}{2}\right)=(3, -1)$ 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 $(3, -1)$, so $h=3$ and $k=-1$. Thus, the standard form of the given circle's equation of the circle is: $(x-3)^2+[y-(-1)]^2=r^2 \\(x-3)^2+(y+1)^2=r^2$ The circle passes through the point $(-1, 3)$. 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: $(x-3)^2+(y+1)^2=r^2 \\(-1-3)^2+(3+1)^2=r^2 \\(-4)^2+4^2=r^2 \\16+16=r^2 \\32=r^2$ Thus, the standard form of the given circle's equation is: $(x-3)^2+(y+1)^2=32$
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