A community for students.
Here's the question you clicked on:
 0 viewing
Lilmike234
 one year ago
What is the center and radius of the circle (x+4)^2+(y2)^2=16
Lilmike234
 one year ago
What is the center and radius of the circle (x+4)^2+(y2)^2=16

This Question is Closed

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0Hint: Rewrite \[\Large (x+4)^2+(y2)^2=16\] as \[\Large (x(4))^2+(y2)^2=4^2\] and compare that to the general form\[\Large (xh)^2+(yk)^2=r^2\]

lilmike234
 one year ago
Best ResponseYou've already chosen the best response.0So the center is (4,2) and the radius is 16

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0for the r what is the square root of 16?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0yes... so our radius is actually 4. our center is in the form of (h,k). I think opposites signs are needed. I have to recheck this one fast one.

lilmike234
 one year ago
Best ResponseYou've already chosen the best response.0So the center is (4,2)

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ah it's like solving for x and solving for y yeah(4,2) is the center... radius is 4

jim_thompson5910
 one year ago
Best ResponseYou've already chosen the best response.0\[\Large (x({\color{red}{4}}))^2+(y{\color{blue}{2}})^2={\color{green}{4}}^2\] \[\Large (x{\color{red}{h}})^2+(y{\color{blue}{k}})^2={\color{green}{r}}^2\]

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0ok... it's (4,2) for center radius is 4 ... whew just double checking. I"m 75% sleepy
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.