A parametrization describes the unit circle if $$x(t)^2 + y(t)^2 = 1$$. For option D, $$x(t)^2 + y(t)^2 = (\cos(t))^2 + (2\sin(t))^2 = \cos^2(t) + 4\sin^2(t)$$, which does not equal 1 for all $$t$$. This parametrization actually describes an ellipse.

The standard counterclockwise parametrization for a circle with radius $$r$$ centered at the origin is $$x(t) = r\cos(t), y(t) = r\sin(t)$$. With a radius of 5, this becomes $$x(t) = 5\cos(t), y(t) = 5\sin(t)$$. At $$t=0$$, the position is $$(5\cos(0), 5\sin(0)) = (5,0)$$, which is the correct starting point.

The path starts at (0,0) and needs to end at (4,6) when $$t=2$$. The parametrization will be of the form $$x(t)=at, y(t)=bt$$. At $$t=2$$, we need $$x(2)=4$$ and $$y(2)=6$$. So, $$a(2)=4 \implies a=2$$, and $$b(2)=6 \implies b=3$$. The correct parametrization is $$x(t)=2t, y(t)=3t$$ for the interval $$0 \le t \le 2$$.

The center is (0, 50) and radius is 40. The period is 2 minutes, so $$P=2 = 2\pi/B$$, which gives $$B=\pi$$. The lowest point is $$(0, 50-40) = (0, 10)$$. Standard counterclockwise motion from the rightmost point is $$x=40\cos(\pi t), y=50+40\sin(\pi t)$$. The lowest point is at an angle of $$-\pi/2$$ from the start. A phase shift gives $$x(t) = 40\cos(\pi t - \pi/2) = 40\sin(\pi t)$$ and $$y(t) = 50 + 40\sin(\pi t - \pi/2) = 50 - 40\cos(\pi t)$$. At $$t=0$$, this gives $$(0, 10)$$, the correct starting point.

The period of the motion is 10 seconds. For a parametrization $$x(t) = r\cos(Bt), y(t) = r\sin(Bt)$$, the period is $$P = 2\pi/|B|$$. We are given $$P=10$$, so $$10 = 2\pi/B$$, which gives $$B = 2\pi/10 = \pi/5$$. The radius is 4. Thus, the correct parametrization is $$x(t) = 4\cos(\frac{\pi}{5}t), y(t) = 4\sin(\frac{\pi}{5}t)$$.

To reverse the direction of a counterclockwise parametrization to clockwise, one common method is to replace $$t$$ with $$-t$$. Starting with $$x(t) = \cos(t)$$ and $$y(t) = \sin(t)$$, this gives $$x(-t) = \cos(-t) = \cos(t)$$ and $$y(-t) = \sin(-t) = -\sin(t)$$. This traces the circle clockwise and starts at $$(1,0)$$ when $$t=0$$. Another method is to negate the sine component.

A line segment from $$(x_1, y_1)$$ to $$(x_2, y_2)$$ for $$0 \le t \le 1$$ can be parametrized as $$x(t) = x_1 + (x_2 - x_1)t$$ and $$y(t) = y_1 + (y_2 - y_1)t$$. Here, $$(x_1, y_1) = (1, 2)$$ and $$(x_2, y_2) = (5, 8)$$. So, $$x(t) = 1 + (5 - 1)t = 1 + 4t$$ and $$y(t) = 2 + (8 - 2)t = 2 + 6t$$.

This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (r). For this question, the circle has center (-2,1) and radius 6, while the line passes through (4,1) with slope 2. Choice A is correct because substituting x=4+r and y=1+2r into (x+2)²+(y-1)²=36 gives r=0 and r=-4, yielding intersection points (4,1) and (0,-7). Choice C is incorrect as it finds one correct point but miscalculates the second intersection by using the wrong parameter value. To help students: Practice solving quadratic equations that arise from circle-line intersections. Verify solutions by checking that both points satisfy the original circle equation.

This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (u). For this question, the circle has center (1,2) and radius 4, while the line passes through (5,6) with slope -1. Choice A is correct because substituting x=5+u and y=6-u into the circle equation (x-1)²+(y-2)²=16 gives u=0 and u=4, yielding intersection points (5,6) and (9,2). Choice C is incorrect due to an algebraic error in solving the quadratic equation, resulting in wrong parameter values. To help students: Emphasize careful algebraic manipulation when substituting parametric equations. Practice converting between parametric and Cartesian forms, and verify solutions by substituting back into both original equations.

This question tests AP Precalculus skills: understanding parametrically defined circles and lines (focus on algebraic and geometric interpretation). Parametric equations represent geometric figures by defining coordinates as functions of a parameter, often time (t) or distance (k). For this question, the circle has center (-1,4) and radius 3, while the line passes through (2,4) with slope 2. Choice A is correct because substituting x=2+k and y=4+2k into (x+1)²+(y-4)²=9 gives k=0 and k=-2, yielding intersection points (2,4) and (0,0). Choice C is incorrect as it finds one correct point but uses the wrong parameter value for the second intersection, resulting in a point outside the circle. To help students: Develop systematic approaches to solving circle-line intersection problems. Practice interpreting parameter values geometrically to understand what they represent on the line.