Fractions and Decimals

Discrete Mathematics

Differential Equations

Solved on Feb 22, 2024

Find the center of the circle with equation $(x+3)^{2} + y^{2} = 4$ . Provide the simplified ordered pair.

STEP 1

Assumptions
1. We are given the equation of a circle in the form $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle and $r$ is the radius.
2. We need to identify the center of the circle from the given equation $(x+3)^2 + y^2 = 4$ .

STEP 2

Identify the standard form of the equation of a circle.
The standard form of the equation of a circle is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center of the circle and $r$ is the radius.

STEP 3

Compare the given equation with the standard form to identify the center.
The given equation is $(x+3)^2 + y^2 = 4$ . This can be compared to the standard form $(x-h)^2 + (y-k)^2 = r^2$ .

STEP 4

Identify the values of $h$ and $k$ from the given equation.
In the given equation, $x$ is being added to 3, so $h = -3$ . There is no addition or subtraction with $y$ , so $k = 0$ .

STEP 5

Write the center of the circle as an ordered pair.
The center $(h,k)$ is $(-3,0)$ .
The center of the circle is $(-3,0)$ .

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