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x^(2)+y^(2)-12 x-10 y+36=0
What is the length of the radius of the circle whose equation is shown?

$x^{2}+y^{2}-12 x-10 y+36=0$ $\newline$ What is the length of the radius 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}-12 x-10 y+36=0

\newline

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

Complete the Square: First, we need to complete the square for both $x$ and $y$ terms. $\newline$ For $x$ : $x^2 - 12x$ . Half of $-12$ is $-6$ , square it to get $36$ . Add and subtract $36$ inside the equation. $\newline$ For $y$ : $y^2 - 10y$ . Half of $y$ $0$ is $y$ $1$ , square it to get $y$ $2$ . Add and subtract $y$ $2$ inside the equation.
Rewrite Equation: Rewrite the equation including the added terms for completing the square: $x^2 - 12x + 36 + y^2 - 10y + 25 - 36 - 25 + 36 = 0$
Group Perfect Squares: Group the terms to form perfect squares: $\newline$ $(x^2 - 12x + 36) + (y^2 - 10y + 25) = 36 + 25 - 36$
Simplify Right Side: Simplify the right side of the equation: $36 + 25 - 36 = 25$
Write Left Side: Write the left side as perfect squares: $\newline$ $(x - 6)^2 + (y - 5)^2 = 25$
Standard Form for Circle: The equation $(x - 6)^2 + (y - 5)^2 = 25$ is now in standard form for a circle, $(x - h)^2 + (y - k)^2 = r^2$ , where $(h,k)$ is the center and $r$ is the radius.
Identify Radius: Identify the radius of the circle: $\newline$ Since the right side of the equation equals $25$ , and in the form $r^2$ , $r^2 = 25$ .
Solve for r: Solve for r: $\newline$ $r = \sqrt{25}$ $\newline$ $r = 5$

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