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What is the center of the circle x2+y2−19y+26=0x^2+y^2-19y+26=0? \newlineSimplify any fractions. \newline(□,□)(\square, \square)

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Find properties of circles from equations in general formright chevron icon

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Q. What is the center of the circle x2+y2−19y+26=0x^2+y^2-19y+26=0? \newlineSimplify any fractions. \newline(□,□)(\square, \square)
  1. Complete the square for yy: To find the center of the circle, we need to complete the square for both xx and yy. The given equation is x2+y2−19y+26=0x^2 + y^2 - 19y + 26 = 0. We notice that the xx terms are already a perfect square, so we only need to complete the square for the yy terms.
  2. Move constant term: First, we move the constant term to the other side of the equation: x2+y2–19y=−26x^2 + y^2 – 19y = -26.
  3. Add half of coefficient: Next, we complete the square for the y terms. To do this, we take half of the coefficient of yy, which is −192-\frac{19}{2}, square it, which is (−192)2=3614\left(-\frac{19}{2}\right)^2 = \frac{361}{4}, and add it to both sides of the equation.
  4. Simplify the equation: The equation becomes x2+y2−19y+3614=−26+3614x^2 + y^2 - 19y + \frac{361}{4} = -26 + \frac{361}{4}. Simplify the right side of the equation by converting −26-26 to a fraction with the same denominator as 3614\frac{361}{4}, which is −1044-\frac{104}{4}. So, the equation is now x2+y2−19y+3614=−1044+3614x^2 + y^2 - 19y + \frac{361}{4} = -\frac{104}{4} + \frac{361}{4}.
  5. Rewrite yy terms: Simplify the right side of the equation: −1044+3614=2574-\frac{104}{4} + \frac{361}{4} = \frac{257}{4}. Now the equation is x2+y2−19y+3614=2574x^2 + y^2 - 19y + \frac{361}{4} = \frac{257}{4}.
  6. Identify center of circle: We can now rewrite the y terms as a perfect square: x2+(y−192)2=2574x^2 + (y - \frac{19}{2})^2 = \frac{257}{4}. This shows that the circle is in the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center of the circle and rr is the radius.
  7. Identify center of circle: We can now rewrite the y terms as a perfect square: x2+(y−192)2=2574x^2 + (y - \frac{19}{2})^2 = \frac{257}{4}. This shows that the circle is in the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center of the circle and rr is the radius.From the equation x2+(y−192)2=2574x^2 + (y - \frac{19}{2})^2 = \frac{257}{4}, we can see that h=0h = 0 and k=192k = \frac{19}{2}. Therefore, the center of the circle is (0,192)(0, \frac{19}{2}).

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