College Algebra (10th Edition)

Published by Pearson
ISBN 10: 0321979478
ISBN 13: 978-0-32197-947-6

Chapter 2 - Section 2.4 - Circles - 2.4 Assess Your Understanding - Page 186: 43



Work Step by Step

With its center at $(-1, 3)$, the tentative equation of the circle is: $(x-h)^2+(y-k)^2=r^2 \\(x-(-1))^2+(y-3)^2=r^2 \\(x+1)^2+(y-3)^2=r^2$ The circle is tangent to the line $y=2$. This means that the point on the line $y=2$ that is directly under the center is a point on the circle. This point is $(-1, 2)$. (refer to the attached image below) Solve for the value of $r^2$ by substituting the x and y values of the point $(-1, 2)$ into the tentative equation above to obtain: $(x+1)^2+(y-3)^2=r^2 \\(-1+1)^2+(2-3)^2=r^2 \\0^2+(-1)^2=r^2 \\0+1=r^2 \\1=r^2$ Therefore, the equation of the circle is: $(x+1)^2+(y-3)^2=1$
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