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Completing the square is a useful algebraic process to rewrite quadratic expressions into a form that reveals important characteristics, such as the vertex of a parabola or the center of a sphere.
To complete the square, particularly in the context of a sphere's equation, we focus on a specific variable term, like the linear term in a quadratic expression.
Here’s a simple breakdown:

• Isolate the quadratic and linear terms of the variable that you are completing the square for.
• Take the coefficient of the linear term, divide it by two, and square the result.
• Add and subtract this squared value within the expression to preserve the equality.
• Rewrite the expression as a perfect square trinomial and a constant.

In our case, with the z-terms given as \(z^2 - 6z\), we take -6, divide by 2 to get -3, and square it to get 9.
Thus, we add and subtract 9, rearranging the expression into the form \((z-3)^2 - 9\).
This turns a simple quadratic expression into a format that is easy to work with when identifying the sphere's center and radius.

To find the center of a sphere from its equation in the standard form, you need the equation structured as:\((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\).Here,

• \(a\), \(b\), and \(c\) are the coordinates of the sphere's center \((a, b, c)\).

For the given exercise, we transformed the equation into \(x^2 + y^2 + (z-3)^2 = 25\).
This setup directly showcases that:- The x-term doesn't translate from completion, indicating its coefficient is 0 (hence no shift from zero for x).- Similarly, the y-term's coefficient is also 0.- The z-term is in the form \((z-3)^2\), meaning a shift of 3 units along the z-axis.So, the center of the sphere is at the coordinates \((0, 0, 3)\). Such a center placement represents a sphere centrally positioned on a 3D axis, conveniently simplified and intuitive for spatial understanding.

The radius of a sphere is a fundamental measure that tells us the distance from the sphere's center to any point on its surface.
It's derived easily from the equation of a sphere, particularly when the equation is in its standard form:\((x-a)^2 + (y-b)^2 + (z-c)^2 = r^2\).Here, \(r^2\) represents the square of the radius, \(r\).
In the step-by-step solution, we work with the expression \(x^2 + y^2 + (z-3)^2 = 25\).
The number 25 here tells us that \(r^2 = 25\).
Taking the square root of this result provides the actual length of the radius:\(r = \sqrt{25} = 5\).Having the radius allows us to fully visualize the sphere's size and perceive its reach through three-dimensional space.
Whether for collision detection in graphics, geometrical interpretation, or physics simulations, knowing the radius alongside the center gives a clear understanding of the sphere’s location and size.