Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 1 - Section 1.9 - The Coordinate Plane; Graphs of Equations; Circles - 1.9 Exercises - Page 104: 94

Answer

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

Work Step by Step

Endpoints of a diameter are $P(-1,3)$ and $Q(7,-5)$ The equation of a circle is $(x-h)^{2}+(y-k)^{2}=r^{2}$, where $(h,k)$ is the center of the circle and $r$ is its radius. Use the formula for the distance between two points to find the distance between the points given, which represents the diameter of the circle. The formula is $d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$ $d=\sqrt{(7+1)^{2}+(-5-3)^{2}}=\sqrt{(8)^{2}+(-8)^{2}}=\sqrt{64+64}=...$ $...=\sqrt{128}=8\sqrt{2}$ Since the radius of a circle is half its diameter, then the radius of the circle is: $r=\dfrac{8\sqrt{2}}{2}=4\sqrt{2}$ Find the midpoint of the diameter $PQ$ to obtain the center of the circle: $\Big(\dfrac{-1+7}{2},\dfrac{3-5}{2}\Big)=\Big(\dfrac{6}{2},\dfrac{-2}{2}\Big)=(3,-1)$ Since now both the center and the radius of the circle are known, substitute them into the equation of a circle formula: $(x-h)^{2}+(y-k)^{2}=r^{2}$ $(x-3)^{2}+[y-(-1)]^{2}=(4\sqrt{2})^{2}$ $(x-3)^{2}+(y+1)^{2}=16(2)$ $(x-3)^{2}+(y+1)^{2}=32$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.