To find the equation of a circle given its center and radius, we use the standard form of the equation of a circle:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
Where:
- [tex]\((h, k)\)[/tex] is the center of the circle.
- [tex]\(r\)[/tex] is the radius of the circle.
Given in the question:
- The center of the circle [tex]\((h, k)\)[/tex] is [tex]\((2, -3)\)[/tex].
- The radius [tex]\(r\)[/tex] is 3.
Now, we substitute the given center and radius values into the standard form equation:
1. Replace [tex]\(h\)[/tex] with 2:
[tex]\[
(x - 2)^2
\][/tex]
2. Replace [tex]\(k\)[/tex] with -3. Note that subtracting -3 is the same as adding 3:
[tex]\[
(y - (-3))^2 \implies (y + 3)^2
\][/tex]
3. Replace [tex]\(r\)[/tex] with 3 and remember to square it:
[tex]\[
r^2 = 3^2 = 9
\][/tex]
So, substituting these into the standard form, we get the equation of the circle:
[tex]\[
(x - 2)^2 + (y + 3)^2 = 9
\][/tex]
Therefore, the equation of the circle is:
[tex]\[
(x - 2)^2 + (y + 3)^2 = 9
\][/tex]
Examining the choices provided:
- A. [tex]\((x-2)^2-(y+3)^2=9\)[/tex]
- B. [tex]\((x-2)^2+(y+3)^2=3\)[/tex]
- C. [tex]\((x+2)^2+(y-3)^2=9\)[/tex]
- D. [tex]\((x-2)^2+(y+3)^2=9\)[/tex]
The correct choice is D. [tex]\((x-2)^2+(y+3)^2=9\)[/tex].
So, the correct answer to the question is:
[tex]\[
D. (x-2)^2+(y+3)^2=9
\][/tex]
