$x^{2}+y^{2}+6 x-4 y=3$ $\newline$ A circle in the $x y$ -plane has the equation shown. What is the $y$ coordinate of the center of the circle?

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Algebra 2

Find properties of circles from equations in general form

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x^{2}+y^{2}+6 x-4 y=3

\newline

A circle in the

x y

-plane has the equation shown. What is the

y

coordinate of the center of the circle?

Rewriting the equation: To find the center of the circle, we need to rewrite the equation in the standard form of a circle, which is $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle. We will complete the square for both $x$ and $y$ .
Grouping the terms: First, we group the x terms and the y terms together: $(x^2 + 6x) + (y^2 - 4y) = 3$ .
Completing the square for $x$ terms: To complete the square for the $x$ terms, we take half of the coefficient of $x$ , which is $\frac{$ $6$ }{ $2$ } = $3$ , and square it, adding $3$ ^ $2$ = $9$ to both sides of the equation.
Completing the square for $y$ terms: Similarly, to complete the square for the $y$ terms, we take half of the coefficient of $y$ , which is $-\frac{$ $4$ }{ $2$ } = $-2$ , and square it, adding $($ $-2$ )^ $2$ = $4$ to both sides of the equation.
Adding constants to both sides: Now we add $9$ and $4$ to both sides of the equation to maintain equality: $(x^2 + 6x + 9) + (y^2 - 4y + 4) = 3 + 9 + 4$ .
Simplifying the right side: Simplify the right side of the equation: $3 + 9 + 4 = 16$ .
Standard form of the equation: The equation now looks like this: $(x^2 + 6x + 9) + (y^2 - 4y + 4) = 16$ .
Finding the center of the circle: We can now rewrite the equation as $(x + 3)^2 + (y - 2)^2 = 16$ , which is in the standard form of a circle.
The y-coordinate of the center: From the standard form $(x + 3)^2 + (y - 2)^2 = 16$ , we can see that the center of the circle is at $(-3, 2)$ .
The y-coordinate of the center: From the standard form $(x + 3)^2 + (y - 2)^2 = 16$ , we can see that the center of the circle is at $(-3, 2)$ .The y-coordinate of the center of the circle is $2$ .