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What is the center and radius of the circle?
(x-2)² + y² = 9
The center of the circlip is ☐
(Type an ordered pair.)
The radius of the circle is
(Simplify your answer.)


Sagot :

To determine the center and radius of the circle given by the equation [tex]\((x - 2)^2 + y^2 = 9\)[/tex], we can compare it to the standard form of the equation of a circle, which is [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex]. Here, [tex]\((h, k)\)[/tex] represents the center of the circle, and [tex]\(r\)[/tex] represents the radius.

1. Identify the center [tex]\((h, k)\)[/tex]:

The given equation is [tex]\((x - 2)^2 + y^2 = 9\)[/tex]. From the standard form [tex]\((x - h)^2 + (y - k)^2 = r^2\)[/tex], we can see that:
- [tex]\(h\)[/tex] is the value that [tex]\(x\)[/tex] is subtracted from inside the parentheses. Here, [tex]\(x\)[/tex] is subtracted from 2, so [tex]\(h = 2\)[/tex].
- [tex]\(k\)[/tex] is the value that [tex]\(y\)[/tex] is subtracted from inside the parentheses. Here, [tex]\(y\)[/tex] is not subtracted from anything, so [tex]\(k = 0\)[/tex].
Therefore, the center of the circle is [tex]\((2, 0)\)[/tex].

2. Identify the radius [tex]\(r\)[/tex]:

The right-hand side of the equation is [tex]\(r^2\)[/tex]. In the given equation, this value is 9. To find the radius [tex]\(r\)[/tex], take the square root of 9:
[tex]\[ r = \sqrt{9} = 3 \][/tex]
Therefore, the radius of the circle is 3.

So, the answers are:

- The center of the circle is [tex]\((2, 0)\)[/tex].
- The radius of the circle is 3.
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