Answer
$(x-2)^2 + (y+5)^2=6^2$
center: $(2,-5)$ $\qquad$ radius: $6$
Work Step by Step
$x^2 -4x+y^2+10y-7=0$
$(x^2-4x \textbf{+4}) + (y^2+10y \textbf{+25})=7 \textbf{+4+25}$
$(x-2)^2 + (y+5)^2=36$
center: $(2,-5)$ $\qquad$ radius: $6$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 3 - Linear and Quadratic Functions - 3.1 Properties of Linear Functions and Linear Models - 3.1 Assess Your Understanding - Page 129: 57
Update this answer!
You can help us out by revising, improving and updating this answer.
Update this answerAfter 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.
