To complete the parametrizations for the circle, we need to recall the general equation of a circle in parametric form. The equation of a circle with center $(h, k)$ and radius $r$ is:

$x = h + r \cos(t)$ $y = k + r \sin(t)$

In this case, the circle is centered at $(2, -9)$ and has a radius of $6$. Therefore, the equation of the circle is:

$x = 2 + 6 \cos(t)$ $y = -9 + 6 \sin(t)$

Now, let's look at the parametrizations given:

Part (a)

You are given: $x = 2 + 6 \cos(t)$ The corresponding $y$-coordinate is already provided: $y = -9 + 6 \sin(t)$ So, part (a) is complete.

Part (b)

You are given: $x = 2 + 6 \sin(t)$ For the corresponding $y$-coordinate, we know it still follows the form of the parametric equation of the circle, so: $y = -9 + 6 \cos(t)$ This is the corresponding parametrization for part (b).

Final Answer:

• (a) $y = -9 + 6 \sin(t)$
• (b) $y = -9 + 6 \cos(t)$

Would you like a more detailed explanation of how to derive these parametrizations? Or do you have any other questions?

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Here are some related questions:

1. How does the direction of traversal (clockwise vs counter-clockwise) affect the parametrization?
2. How do we derive parametric equations from the general circle equation?
3. Can we parametrize an ellipse similarly to a circle? If so, how?
4. What happens to the parametrization if we change the center of the circle?
5. How would you parametrize a circle if the radius were negative?

Tip: Always ensure the parametric equations respect the sign and direction constraints of the problem.