### Video Transcript

Convert the rectangular equation π₯
squared plus π¦ squared equals 25 to parametric form.

Letβs begin by recalling what we
actually know about this equation. We know that a circle whose centre
is at the origin and whose radius is π can be given by the Cartesian equation π₯
squared plus π¦ squared equals π squared. By rewriting our rectangular
equation as π₯ squared plus π¦ squared equals five squared, we see that we have a
circle whose centre is at the origin and whose radius is five units. And so, Iβve sketched that on the
π₯π¦-plane, as shown.

Weβre looking to convert this
equation to parametric form. So we know that a pair of
parametric equations describe the π₯- and π¦-coordinates in terms of a third
parameter, π‘. So letβs pick a general point π₯,
π¦. Iβm going to choose this one in the
first quadrant. We can drop in a right triangle,
whose height is π¦ units and whose width is π₯ units. And then, we can label the included
angle π‘. Since the radius of the circle is
five units, we know that the hypotenuse of our triangle is five. Then, using standard conventions,
we label the sides of our triangle. We have the adjacent; thatβs
π₯. We have the opposite; thatβs
π¦. We have the hypotenuse; thatβs
five.

We also know that in rectangular
trigonometry, sin π is equal to the opposite divided by the hypotenuse and cos π
is equal to the adjacent over the hypotenuse. So we can say that sin π‘ equals π¦
over five and cos π‘ is equal to π₯ over five. We can multiply through by five for
both of our equations. And we find that π¦ is equal to
five sin π‘ and π₯ equals five cos π‘. So for any point in our circle, the
π¦-coordinate is given by five sin of π‘ and the π₯-coordinate is given by five cos
of π‘. Now moving in a counterclockwise
direction from the positive horizontal axis, we see that as π₯ increases from zero,
it generates our corresponding π₯- and π¦-coordinates. Then, our equations are π₯ equals
five cos π‘ and π¦ equals five sin π‘.