What is the equation of a circle with its center at (−6,−3) and a radius of 12?
A: (x−6)2+(y−3)2=12
B: (x−6)2+(y−3)2=144
C: (x+6)2+(y+3)2=144
D: (x+6)2+(y+3)2=12
E: (x+6)2+(y+3)2=24
I believe that it is C
Also, the 2's after the parenthesis are supposed the be exponents
I believe that you're correct
To determine the equation of a circle with its center at (h, k) and a radius of r, we use the standard form equation:
(x - h)^2 + (y - k)^2 = r^2
In this case, the center of the circle is (-6, -3), and the radius is 12. Plugging these values into the equation, we get:
(x - (-6))^2 + (y - (-3))^2 = 12^2
(x + 6)^2 + (y + 3)^2 = 144
Therefore, the correct answer is B: (x − 6)^2 + (y − 3)^2 = 144.