x^(2)+y^(2)-6x+14 y=6 What is the length of the diameter of the circle whose equation is shown?

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Equation in Standard Form: The given equation is x^(2) + y^(2) - 6x + 14y = 6. To find the length of the diameter, we first need to write 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 and r is the radius.Completing the Square: We complete the square for the x-terms and the y-terms in the equation. For the x-terms, we take the coefficient of x, which is -6, divide it by 2 to get -3, and then square it to get 9. We add and subtract this value inside the equation. For the y-terms, we take the coefficient of y, which is 14, divide it by 2 to get 7, and then square it to get 49. We add and subtract this value inside the equation as well.Rewriting the Equation: The equation becomes x^(2) - 6x + 9 + y^(2) + 14y + 49 = 6 + 9 + 49. We added 9 and 49 to both sides of the equation to keep the equation balanced.Finding the Radius: Now, we can rewrite the equation as (x - 3)^(2) + (y + 7)^(2) = 64. This is the standard form of the equation of a circle, where the center is at (h, k) = (3, -7) and the radius squared, r^(2), is 64.Calculating the Diameter: To find the radius r, we take the square root of 64, which gives us r = 8.The diameter d of a circle is twice the radius. Therefore, the diameter is d = 2 * r.Substituting the value of the radius we found, we get d = 2 * 8 = 16.

$x^{2}+y^{2}-6 x+14 y=6$ $\newline$ What is the length of the diameter of the circle whose equation is shown?

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x2+y2−6x+14y=6 x^{2}+y^{2}-6 x+14 y=6 x2+y2−6x+14y=6\newlineWhat is the length of the diameter of the circle whose equation is shown?

$x^{2}+y^{2}-6 x+14 y=6$ $\newline$ What is the length of the diameter of the circle whose equation is shown?