To determine the center and radius of the circle, we start by comparing the given equation of the circle [tex]\((x-2)^2 + (y+3)^2 = 36\)[/tex] with the general form of a circle's equation [tex]\((x-h)^2 + (y-k)^2 = r^2\)[/tex]. The general form helps us identify the center [tex]\((h, k)\)[/tex] and the square of the radius [tex]\(r^2\)[/tex].
Let's break it down step by step:
1. Identify the Center:
- The equation given is [tex]\((x-2)^2 + (y+3)^2 = 36\)[/tex].
- Here, [tex]\((h, k)\)[/tex] are the coordinates that shift the circle from the origin.
- From the comparison, we see that [tex]\(h = 2\)[/tex] and [tex]\(k = -3\)[/tex].
Therefore, the center of the circle is:
[tex]\[
(h, k) = (2, -3)
\][/tex]
2. Determine the Radius:
- The right-hand side of the equation is the radius squared, which is [tex]\(36\)[/tex].
- To find the radius [tex]\(r\)[/tex], we take the square root of [tex]\(36\)[/tex]:
[tex]\[
r = \sqrt{36} = 6
\][/tex]
Thus, the center of the circle is [tex]\((2, -3)\)[/tex] and the radius is [tex]\(6\)[/tex].
Correct Answer:
- Center at [tex]\((2,-3)\)[/tex] and radius [tex]\(r = 6\)[/tex]
Option with:
- Center at [tex]\((2,-3)\)[/tex] ; [tex]\(r=6\)[/tex]
