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The upper half of a circle is visualized by considering the positive portion of the vertical axis, specifically the positive y-values above the x-axis. For a circle described by the equation \( (x+3)^2 + y^2 = 64 \), the center of the circle is at the point \((-3, 0)\) with a radius of 8 units. This base circle includes all the points in its entirety, both above and below the x-axis.
To isolate the upper half, you need to solve for \( y \), extracting only the positive values. The process involves manipulating the circle's equation to \( y = \sqrt{64 - (x+3)^2} \). This equation represents the segment of points in the upper semicircle where the y-values are positive, scooping up all points found above the x-axis.
The graphical representation of this form is the semicircular arc on top, which stretches from the leftmost point at \((-11, 0)\) to the rightmost point at \((5, 0)\) along the x-axis, because the endpoints of a full circle lie symmetrically relative to its center.

For the lower half of the circle, we focus on the negative portion of the vertical axis, meaning the negative y-values. The lower semicircle is a reflection beneath the x-axis of what you see with the upper half.
When dealing with the equation of the circle, \( (x+3)^2 + y^2 = 64 \), solving for \( y \) involves identifying \( y = -\sqrt{64 - (x+3)^2} \). This equation caters to the negative y-values, portraying the lower hemisphere of the circle's form.
This lower boundary sweeps from \((-11, 0)\) to \((5, 0)\), just like its upper counterpart, but below the x-axis, forming a semicircular path. This segment is essential for understanding problems involving complete circular geometry but confined to the lower half.

The right half of a circle refers to the portion extending along the positive side of the horizontal axis rather than related to y-values. It embodies all points where the x-coordinate is greater than or equal to the circle's central x-coordinate. For our equation \( (x+3)^2 + y^2 = 64 \), the origin is at \((-3, 0)\), meaning the right half includes \( x \geq -3 \).
To express this mathematically, solve for \( x \) with the equation \( x = \sqrt{64 - y^2} - 3 \), capturing the positive component of x-values. This sets a scene where the circle's right semicircle arcs around the center, sweeping from top to bottom of its visible hemisphere, crossing the full span of y-values, but keeping to the right of \( x = -3 \).
It's a vital segment for understanding problems that restrict movement or placement within one side of a circular boundary.

Similarly to its right counterpart, the left half of a circle encompasses the entire negative x-value side relative to the circle's center. With the equation \((x+3)^2 + y^2 = 64\), since the circle orbits around \((-3, 0)\), the left circle half is described as \( x \leq -3 \).
This involves altering the circle's formula to \( x = -\sqrt{64 - y^2} - 3 \) for the negative root accounting for negative x-values. Therefore, this semicircle sweeps across all possible y-values but hovers over the left of the circle’s central axis on the horizontal plane.
Understanding this segment is beneficial for solving constrained-focal circular boundary problems needing emphasis on expansion towards negative x-direction, particularly when intersecting with other shapes or lines.