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What is the diameter of the circle in the $xy$-plane with equation $\\left(x-5\\right)^{2}+\\left(y-3\\right)^{2}=16$?

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What is the diameter of the circle in the $xy$-plane with equation $\\left(x-5\\right)^{2}+\\left(y-3\\right)^{2}=16$?

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Choice B is correct. The standard form of an equation of a circle in the $xy$-plane is $\\left(x-h\\right)^{2}+\\left(y-k\\right)^{2}=r^{2}$, where the coordinates of the center of the circle are $\\left(h,k\\right)$ and the length of the radius of the circle is $r$. For the circle in the $xy$-plane with equation $\\left(x-5\\right)^{2}+\\left(y-3\\right)^{2}=16$, it follows that $r^{2}=16$. Taking the square root of both sides of this equation yields $r=4$ or $r=-4$. Because $r$ represents the length of the radius of the circle and this length must be positive, $r=4$. Therefore, the radius of the circle is 4. The diameter of a circle is twice the length of the radius of the circle. Thus, $2(4)$ yields 8. Therefore, the diameter of the circle is 8.

Choice A is incorrect. This is the radius of the circle.

Choice C is incorrect. This is the square of the radius of the circle.

Choice D is incorrect and may result from conceptual or calculation errors.

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What is the diameter of the circle in the $xy$-plane with equation $\\left(x-5\\right)^{2}+\\left(y-3\\right)^{2}=16$?

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What is the diameter of the circle in the $xy$-plane with equation $\left(x-5\right)^{2}+\left(y-3\right)^{2}=16$?