To determine the equation of the circle with a radius of 2 units and the same center as a circle whose equation is given by [tex]\( x^2 + y^2 - 8x - 6y + 24 = 0 \)[/tex], we follow these steps:
1. Start with the given equation of the circle: [tex]\( x^2 + y^2 - 8x - 6y + 24 = 0 \)[/tex].
2. We need to find the center of this circle. To do this, we will complete the square for both the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms.
3. Rewrite the equation by grouping the [tex]\( x \)[/tex] and [tex]\( y \)[/tex] terms:
[tex]\[
x^2 - 8x + y^2 - 6y + 24 = 0
\][/tex]
4. Complete the square for the [tex]\( x \)[/tex] terms:
[tex]\[
x^2 - 8x \rightarrow (x - 4)^2 - 16
\][/tex]
5. Complete the square for the [tex]\( y \)[/tex] terms:
[tex]\[
y^2 - 6y \rightarrow (y - 3)^2 - 9
\][/tex]
6. Substitute the completed squares back into the equation:
[tex]\[
(x - 4)^2 - 16 + (y - 3)^2 - 9 + 24 = 0
\][/tex]
7. Simplify the equation:
[tex]\[
(x - 4)^2 + (y - 3)^2 - 25 + 24 = 0
\][/tex]
[tex]\[
(x - 4)^2 + (y - 3)^2 - 1 = 0
\][/tex]
[tex]\[
(x - 4)^2 + (y - 3)^2 = 1
\][/tex]
8. The center of the original circle is [tex]\((4, 3)\)[/tex].
9. Now, given that we need to find an equation of a new circle with a radius of 2 units and the same center [tex]\((4, 3)\)[/tex], we start with the standard form of a circle equation:
[tex]\[
(x - h)^2 + (y - k)^2 = r^2
\][/tex]
where [tex]\((h, k)\)[/tex] is the center and [tex]\(r\)[/tex] is the radius.
10. Substitute [tex]\(h = 4\)[/tex], [tex]\(k = 3\)[/tex], and [tex]\(r = 2\)[/tex]:
[tex]\[
(x - 4)^2 + (y - 3)^2 = 2^2
\][/tex]
11. Simplify:
[tex]\[
(x - 4)^2 + (y - 3)^2 = 4
\][/tex]
Therefore, the equation that represents the new circle is:
[tex]\[
(x - 4)^2 + (y - 3)^2 = 4
\][/tex]
So the correct choice among the given options is:
[tex]\[
(x - 4)^2 + (y - 3)^2 = 2^2
\][/tex]
