Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 0 - Before Calculus - 0.3 Families Of Function - Exercises Set 0.2 - Page 35: 5

Answer

$$y=\pm \frac{\sqrt{b^2-9}}{3}+b$$

Work Step by Step

The general equation of a line is $y=mx+b$, and the equation of the circle with center at the origin and radius $3$ is $x^2+y^2=9$. If we want that the line is tangent to the circle, we must have $$x^2+(mx+b)^2=9 \quad \Rightarrow \quad (1+m^2)x^2+2mbx+b^2-9=0 \\ \Delta =(2mb)^2-4(1+m^2)(b^2-9)=0 \quad \Rightarrow \quad b^2-9-9m^2=0 \quad \Rightarrow \quad m=\pm \frac{\sqrt{b^2-9}}{3}.$$Thus, an equation for the family of lines tangent to the circle with center at the origin and radius $3$ is $$y=\pm \frac{\sqrt{b^2-9}}{3}+b \, .$$
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