Thank you for your help! The question is what is the center and radius of the circle. (x+2)^2+y^2=64. Thank you so much.
If your equation is written in the form
(x-h)^2 + (y-k)^2 = r^2 , the centre is (h,k) and the radius is r
Your equation is written in that form and the centre and radius can be seen as is.
Let me know what you got.
Yup. That is what I got reiny. :) Thanks again.
ok, so my final answer is h=2, k=0, r^2=64, then r=8, is this right?
To determine the center and radius of a circle, we can rewrite the equation in the standard form of a circle, which is in the form (x - h)^2 + (y - k)^2 = r^2.
In the given equation (x + 2)^2 + y^2 = 64, we see that the square terms are already in the correct form. Now, we need to isolate the constant term on the right side of the equation.
To do that, we subtract 64 from both sides of the equation:
(x + 2)^2 + y^2 - 64 = 0
Now, we have our equation in the standard form of a circle. By comparing it to (x - h)^2 + (y - k)^2 = r^2, we can determine the center and radius of the circle.
The center of the circle is (-h, -k), so in this case, the center is (-2, 0).
The radius of the circle is the square root of r^2, so in this case, the radius is √64, which is 8.
Therefore, the center of the circle is (-2, 0) and the radius is 8.