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How do you find the parametric equations of a circle?

Answer 1

We'll start with the parametric equations for a circle:

#y = rsin t#
#x = rcos t#

where #t# is the parameter and #r# is the radius.

If you know that the implicit equation for a circle in Cartesian coordinates is #x^2 + y^2 = r^2# then with a little substitution you can prove that the parametric equations above are exactly the same thing.

We will take the equation for #x#, and solve for #t# in terms of #x#:

#x/r = cos t#
#t = arccos (x/r)#

Now substitute into the equation for #y#. This eliminates the parameter #t# and gives us an equation with only #x# and #y#.

#y = rsin arccos(x/r)#

#sin arccos(x/r)# is equal to #sqrt(r^2 - x^2)/r#. This is apparent if one sketches a right triangle, letting #theta = arccos(x/r)#:

Thus, #sin theta = sqrt(r^2 - x^2)/r#. So now we have

#y = r*sqrt(r^2 - x^2)/r#

This simplifies to

#y = sqrt(r^2 - x^2)#

If we square this entire deal and solve for #r#, we get:

#r^2 = x^2 + y^2#

which is precisely the equation for a circle in Cartesian coordinates.

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Answer 2

Christopher Allers

To find the parametric equations of a circle, you can use the parametric form:

$x = x_c + r \cdot \cos(\theta)$
$y = y_c + r \cdot \sin(\theta)$

where:
$x_c$ and $y_c$ are the coordinates of the center of the circle,
$r$ is the radius of the circle, and
$\theta$ is the parameter, typically ranging from $0$ to $2\pi$ for a full circle.

By varying $\theta$ from $0$ to $2\pi$ , you can generate points along the circumference of the circle.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.