SOLUTION: 7. What is the standard equation of the circle with radius 5, and center (-5,-1)? A (x-5)^2 + (y-1)^2 = 25 B (x+5)^2 + (y+1)^2 = 25 C (x+5)^2 + (y+5)^2 = 5 D A circle canno

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 7. What is the standard equation of the circle with radius 5, and center (-5,-1)? A (x-5)^2 + (y-1)^2 = 25 B (x+5)^2 + (y+1)^2 = 25 C (x+5)^2 + (y+5)^2 = 5 D A circle canno      Log On


   

Question 162018: 7. What is the standard equation of the circle with radius 5, and center (-5,-1)?
A (x-5)^2 + (y-1)^2 = 25
B (x+5)^2 + (y+1)^2 = 25
C (x+5)^2 + (y+5)^2 = 5
D A circle cannot have a center at (-5,-1).
E (x-5)^2 - (y-1)^2 = 25
F (x+5)^2 - (y+1)^2 = 25

I used the formula for a circle and plugged the information in, and I got F for the right answer. But choice D is confusing me and i wanted to make sure F really is correct. Thank you!
Found 2 solutions by jim_thompson5910, scott8148:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Since the center is (-5,-1) this means that h=-5 and k=-1. Remember, the center is at (h,k). Also, since the radius is 5, this tells us that r=5


%28x-h%29%5E2%2B%28y-k%29%5E2=r%5E2 Start with the general equation of any circle


%28x-%28-5%29%29%5E2%2B%28y-%28-1%29%29%5E2=5%5E2 Plug in h=-5, k=-1, and r=5


%28x%2B5%29%5E2%2B%28y%2B1%29%5E2=5%5E2 Rewrite x-%28-5%29 as x%2B5. Rewrite y-%28-1%29 as y%2B1


%28x%2B5%29%5E2%2B%28y%2B1%29%5E2=25 Square 5 to get 25


So the standard equation of the circle with a radius of 5 and the center (-5,-1) is %28x%2B5%29%5E2%2B%28y%2B1%29%5E2=25 which means that the answer is B

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
for a circle of radius r, centered at (h,k); the general equation is
__ (x-h)^2 + (y-k)^2 = r^2

looks like B to me...