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Find the center and radius of a circle in steps with answer being center being $(8, 0)$ and radius $(2 \sqrt{3})$ the problem is $(x-8)^2+y^2=12$

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Find the magnitude of a three-dimensional vector

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Q. Find the center and radius of a circle in steps with answer being center being

(8, 0)

and radius

(2 \sqrt{3})

the problem is

(x-8)^2+y^2=12

Circle Equation Standard Form: The equation of a circle in standard form is $(x-h)^2 + (y-k)^2 = r^2$ , where $(h,k)$ is the center and $r$ is the radius.
Given Equation Comparison: Given equation: x-8)^2 + y^2 = 12\. Compare it with the standard form to find the center \$h,k.
Center Calculation: The center is $(h,k) = (8,0)$ because $h=8$ and $k=0$ .
Radius Calculation: Now, find the radius by taking the square root of $12$ , which is the value on the right side of the equation.
Radius Calculation: Now, find the radius by taking the square root of $12$ , which is the value on the right side of the equation.The radius $r$ is $\sqrt{12}$ , which simplifies to $2\sqrt{3}$ .

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