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What is the center of the circle with equation [tex]\((x-3)^2 + (y+3)^2 = 4\)[/tex] in the standard [tex]\((x, y)\)[/tex] coordinate plane?

A. [tex]\((3,3)\)[/tex]

B. [tex]\((3,-3)\)[/tex]

C. [tex]\((\sqrt{3},-\sqrt{3})\)[/tex]

D. [tex]\((-3,3)\)[/tex]

E. [tex]\((- \sqrt{3}, \sqrt{3})\)[/tex]

Sagot :

GinnyAnsweridn
To determine the center of the circle given by the equation [tex]\((x - 3)^2 + (y + 3)^2 = 4\)[/tex], we need to compare it to the standard form of a circle's equation.

The standard form of the equation of a circle is:
[tex]\[ (x - h)^2 + (y - k)^2 = r^2 \][/tex]

Here:
- [tex]\((h, k)\)[/tex] represents the center of the circle.
- [tex]\(r\)[/tex] represents the radius of the circle.

Given the equation of the circle:
[tex]\[ (x - 3)^2 + (y + 3)^2 = 4 \][/tex]

We can compare this to the standard form:
- The term [tex]\((x - 3)^2\)[/tex] indicates that [tex]\(h = 3\)[/tex].
- The term [tex]\((y + 3)^2\)[/tex] can be rewritten as [tex]\((y - (-3))^2\)[/tex], indicating that [tex]\(k = -3\)[/tex].

Therefore, the center of the circle is [tex]\((h, k) = (3, -3)\)[/tex].

So, the answer is:
B. [tex]\((3, -3)\)[/tex]
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