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x^(2)+y^(2)-6x+14 y=6
What 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?

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Find properties of circles from equations in general form

Full solution

Q.

x^{2}+y^{2}-6 x+14 y=6

\newline

What is the length of the diameter of the circle whose equation is shown?

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.
Identifying the Center and the Radius Squared: 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 Radius: To find the radius $r$ , we take the square root of $64$ , which gives us $r = 8$ .
Calculating the Diameter: The diameter $d$ of a circle is twice the radius. $\newline$ Therefore, the diameter is $d = 2 \times r$ . $\newline$ Substituting the value of the radius we found, we get $d = 2 \times 8 = 16$ .

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