Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Problems Plus - Problems - Page 272: 11

Answer

The center of the circle is at the point $(0, \frac{5}{4})$

Work Step by Step

Let $x_0$ be the positive x-coordinate where the circle intersects the parabola. Then the point $(x_0, x_0^2)$ is a point of intersection. The slope of the parabola at this point is $2 x_0$ Since the circle is tangent to the parabola at this point, then the line connecting the point $(x_0,x_0^2)$ to the center of the circle is $-\frac{1}{2x_0}$ Let $b$ be the vertical distance from the point $(x_0, x_0^2)$ and the center of the circle. Then $b = \sqrt{1-x_0^2}$ We can find $x_0$: $\frac{b}{-x_0} = -\frac{1}{2x_0}$ $b = \frac{1}{2}$ $\sqrt{1-x_0^2} = \frac{1}{2}$ $1-x_0^2 = \frac{1}{4}$ $x_0^2 = \frac{3}{4}$ $x_0 = \frac{\sqrt{3}}{2}$ The center of the circle is at the point $(0, x_0^2+b)$ We can find $x_0^2+b$: $x_0^2+b = x_0^2+\sqrt{1-x_0^2}$ $x_0^2+b = (\frac{\sqrt{3}}{2})^2+\sqrt{1-(\frac{\sqrt{3}}{2})^2}$ $x_0^2+b = \frac{3}{4}+\sqrt{\frac{1}{4}}$ $x_0^2+b = \frac{3}{4}+\frac{1}{2}$ $x_0^2+b = \frac{5}{4}$ The center of the circle is at the point $(0, \frac{5}{4})$
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