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: 44

Answer

$(x-4)^2+(y+2)^2=9$

Work Step by Step

With its center at $(4, -2)$, the tentative equation of the circle is: $(x-h)^2+(y-k)^2=r^2 \\(x-4)^2+(y-(-2))^2=r^2 \\(x-4)^2+(y+2)^2=r^2$ The circle is tangent to the line $x=1$. This means that the point on the line $x=1$ that is directly to the left of 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: $(x-4)^2+(y+2)^2=r^2 \\(1-4)^2+(-2+2)^2=r^2 \\(-3)^2+0^2=r^2 \\9+0=r^2 \\9=r^2$ Therefore, the equation of the circle is: $(x-4)^2+(y+2)^2=9$
Small 1511421718
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.