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What is the center of the circle $x^2+y^2-19y+26=0$ ? $\newline$ Simplify any fractions. $\newline$ $(\square, \square)$

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Find properties of circles from equations in general form

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Q. What is the center of the circle

x^2+y^2-19y+26=0

?

\newline

Simplify any fractions.

\newline

(\square, \square)

Rearrange equation: To find the center of the circle, we need to complete the square for both $x$ and $y$ . The given equation is $x^2+y^2–19y+26=0$ . We start by rearranging the terms to group $x$ and $y$ terms together. $\newline$ Rearrange the equation: $\newline$ $x^2 + (y^2 - 19y) + 26 = 0$
Complete square for $y$ : Next, we need to complete the square for the $y$ terms. To do this, we take half of the coefficient of $y$ , which is $-\frac{19}{2}$ , and square it. This value is then added and subtracted inside the parentheses to complete the square. $\newline$ Complete the square for $y$ : $\newline$ $(y^2 - 19y + (\frac{19}{2})^2) - (\frac{19}{2})^2 + 26 = 0$ $\newline$ $(y^2 - 19y + 90.25) - 90.25 + 26 = 0$
Rewrite and simplify: Now we rewrite the equation with the completed square and simplify the constants on the right side of the equation. $\newline$ Rewrite and simplify: $\newline$ $(x^2) + (y - \frac{19}{2})^2 = 90.25 - 26$ $\newline$ $(x^2) + (y - \frac{19}{2})^2 = 64.25$
Identify center: The equation now represents a circle in the standard form $(x - h)^2 + (y - k)^2 = r^2$ , where $(h, k)$ is the center of the circle and $r$ is the radius. From the equation $(x^2) + (y - \frac{19}{2})^2 = 64.25$ , we can see that $h = 0$ and $k = \frac{19}{2}$ . $\newline$ Identify the center: $\newline$ $h = 0$ $\newline$ $k = \frac{19}{2}$
Final answer: The center of the circle is at the point $(h, k)$ . Therefore, the center of the circle defined by the equation $x^2+y^2–19y+26=0$ is (0, rac{19}{2}). $\newline$ Final answer: $\newline$ Center of circle: (0, rac{19}{2})

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