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: 76

Answer

$(x+1)^2+(y-5)^2=130$

Work Step by Step

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 $(-1, 5)$, then $h=-1$ and $k=5$. Thus, the standard form of the given circle's equation of the circle is: $[x-(1)]^2+(y-5)^2=r^2 \\(x+1)^2+(y-5)^2=r^2$ The circle passes through the point $(-4, -6)$. 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: $(x+1)^2+(y-5)^2=r^2 \\(-4+1)^2+(-6-5)^2=r^2 \\(-3)^2+(-11)^2=r^2 \\9+121=r^2 \\130=r^2$ Thus, the standard form of the given circle's equation is: $(x+1)^2+(y-5)^2=130$
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