What is the center of the circle given by the equation [tex]$(x+5)^2+(y-8)^2=1$[/tex]?

A. [tex]$(5,8)$[/tex]

B. [tex][tex]$(-5,-8)$[/tex][/tex]

C. [tex]$(5,-8)$[/tex]

D. [tex]$(-5,8)$[/tex]



Answer :

To determine the center of the circle given by the equation [tex]\((x + 5)^2 + (y - 8)^2 = 1\)[/tex], we need to compare this equation to the standard form of a circle's equation, which is:

[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]

In the standard form, [tex]\((h, k)\)[/tex] represents the center of the circle, and [tex]\(r\)[/tex] is the radius of the circle.

Given the equation:

[tex]\[ (x + 5)^2 + (y - 8)^2 = 1 \][/tex]

we can rewrite the terms inside the parentheses to match the standard form [tex]\((x - h)\)[/tex] and [tex]\((y - k)\)[/tex].

Notice that:

[tex]\[ (x + 5)^2 = (x - (-5))^2 \][/tex]

and

[tex]\[ (y - 8)^2 \][/tex]

Thus, comparing this to the standard form [tex]\((x - h)^2 + (y - k)^2\)[/tex], we see that:

[tex]\[ h = -5, \quad k = 8 \][/tex]

So, the center of the circle is:

[tex]\((h, k) = (-5, 8)\)[/tex]

Therefore, the correct answer is:

D. [tex]\((-5, 8)\)[/tex]