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: 12


$r=\sqrt2$ $(x-1)^2+(y-2)^2=(\sqrt2)^2$. $C=(1,2)$

Work Step by Step

$C$ is the midpoint of the two given points, hence $C=(\frac{0+2}{2},\frac{1+3}{2})=(1,2).$ 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: $r=\sqrt{(2-1)^2+(3-2)^2}=\sqrt{1+1}=\sqrt{2}.$ Hence our equation: $(x-1)^2+(y-2)^2=(\sqrt2)^2$.
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