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: 40

Answer

$(x+3)^2+(y-1)^2=9$

Work Step by Step

RECALL: The standard equation of a circle whose center is at the point $(h, k)$ is: $(x-h)^2+(y-k)^2=r^2$ where $r$ = radius With its center at $(-3, 1)$, the tentative equation of the given circle is: $(x-(-3))^2+(y-1)^2=r^2 \\(x+3)^2+(y-1)^2=r^2$ Since the circle is tangent to the $y$-axis, the point on the y-axis that is directly to the right of the center is a point on the circle. This point is $(0, 1)$. Note that the distance from the center $(-3, 1)$ to the point on the circle $(0, 1)$ is 3 units. Therefore, the standard form of the equation of the given circle is: $(x+3)^2+(y-1)^2=3^2 \\(x+3)^2+(y-1)^2=9$
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