Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 1 - Graphs - 1.4 Circles - 1.4 Assess Your Understanding - Page 38: 41

Answer

$(x+1)^2+(y-3)^2=(\sqrt5)^2$.

Work Step by Step

$C$ is the midpoint of the two given points, hence $C=(\frac{1+(-3)}{2},\frac{4+2}{2})=(-1,3).$ The general equation for a circle with radius $r$ and centre $(h,k)$ is: $(x-h)^2+(y-k)^2=r^2$. The distance formula from $P_1(x_1,y_1)$ to $P_2(x_2,y_2)$ is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Hence here: $r=\sqrt{(-1-1)^2+(3-4)^2}=\sqrt{4+1}=\sqrt{5}.$ Hence our equation: $(x+1)^2+(y-3)^2=(\sqrt5)^2$.
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