S sbsbsbsbsb New member Joined Mar 9, 2010 Messages 16 May 9, 2010 #1 Ok the instructions are to sketch the graph of (x-2)^2 + (y-5)^2 < 9 the center is 2, 5 but is the radius 9 or 3?
Ok the instructions are to sketch the graph of (x-2)^2 + (y-5)^2 < 9 the center is 2, 5 but is the radius 9 or 3?
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 May 9, 2010 #2 \(\displaystyle Here \ is \ the \ graph \ of \ (y-5)^2+(x-2)^2 \ = \ 9\) \(\displaystyle Now, \ what \ do \ you \ have \ to \ do \ to \ the \ said \ graph \ to \ have \ (y-5)^2+(x-2)^2 \ \le \ 9?\) [attachment=0:e52tj3fd]zab.jpg[/attachment:e52tj3fd] Attachments zab.jpg 20 KB · Views: 470
\(\displaystyle Here \ is \ the \ graph \ of \ (y-5)^2+(x-2)^2 \ = \ 9\) \(\displaystyle Now, \ what \ do \ you \ have \ to \ do \ to \ the \ said \ graph \ to \ have \ (y-5)^2+(x-2)^2 \ \le \ 9?\) [attachment=0:e52tj3fd]zab.jpg[/attachment:e52tj3fd]
S sbsbsbsbsb New member Joined Mar 9, 2010 Messages 16 May 9, 2010 #3 you have to shade the inside of the circle since all the points in the circle can work right?