Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Appendix A - Appendix A Exercises - Page 1158: 36

Answer

$$\eqalign{ & {\text{Represents the set of all the points lying on or outside the}} \cr & {\text{circle }}{\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {\sqrt {50} } \right)^2} \cr} $$

Work Step by Step

$$\eqalign{ & {x^2} + {y^2} + 6x + 8y \geqslant 25 \cr & {\text{Group terms}} \cr & \left( {{x^2} + 6x} \right) + \left( {{y^2} + 8y} \right) \geqslant 25 \cr & {\text{Complete the square}} \cr & \left( {{x^2} + 6x + 9} \right) + \left( {{y^2} + 8y + 16} \right) \geqslant 25 + 9 + 16 \cr & {\text{Factor and simplify}} \cr & {\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} \geqslant 50 \cr & {\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} \geqslant {\left( {\sqrt {50} } \right)^2} \cr & {\text{We know that the equation }}{\left( {x - h} \right)^2} + {\left( {y - k} \right)^2} = {r^2}, \cr & {\text{is the standard form equation of a circle centered at }} \cr & \left( {h,k} \right){\text{ with radius }}r,{\text{ and represents all the points of the}} \cr & {\text{circle.}}{\text{Then the equation }}{\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} \geqslant {\left( {\sqrt {50} } \right)^2} \cr & {\text{represents the set of all the points lying on or outside the}} \cr & {\text{circle }}{\left( {x + 3} \right)^2} + {\left( {y + 4} \right)^2} = {\left( {\sqrt {50} } \right)^2.} \cr} $$
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.