What is the equation of a circle with its center at (−6,−3) and a radius of 12?
(x+6)2+(y+3)2=12
(x+6)2+(y+3)2=24
(x+6)2+(y+3)2=144
(x−6)2+(y−3)2=12
(x−6)2+(y−3)2=144
could someone pls help me out with this
im very confused
its (x+6)2+(y+3)2=144, just got it right.
If the centre is (a,b) and the radius is r , the equation would be
(x-a)^2 + (y-b)^2 = r^2
hint, since r = 12, r^2 would be 144, I see only two choices with that.
That sure limits it doesn't it?
recall that the equation of a circle with center at (h,k) and radius r is
(x-h)^2 + (y-k)^2 = r^2
so, you know that r^2 = 12^2 = 144
Now, which is your choice?
Of course! I'd be happy to help you understand how to find the equation of a circle with a given center and radius.
The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h, k) represents the coordinates of the center and r is the radius.
In this case, the center of the circle is (-6, -3) and the radius is 12. Plugging these values into the general equation, we get:
(x - (-6))^2 + (y - (-3))^2 = 12^2
Simplifying:
(x + 6)^2 + (y + 3)^2 = 144
So the correct equation of the circle is (x + 6)^2 + (y + 3)^2 = 144. This represents a circle with its center at (-6, -3) and a radius of 12.
I hope this explanation helps clarify the process for finding the equation of a circle given its center and radius! Let me know if you have any further questions.