# A) Find a parametric equation of the circle with radius 3 centered at (2,1) traced...

## Question:

A) Find a parametric equation of the circle with radius 3 centered at (2,1) traced counterclockwise.

B) Repeat the previous problem in the clockwise three-turn.

C) An ellipse {eq}\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 {/eq} can be parametrized counterclockwise by {eq}x(t) = a \cos t {/eq} and {eq}y(t) = b \sin t {/eq}, {eq}0 \leq t \leq 2 \pi{/eq}. Find a parametric equation of the ellipse {eq}4(x-5 )^2+9(y+2)^2 = 36 {/eq} traced clockwise.

## Parametrization for Conics

A conic, like circle or ellipse can be parametrized with cosine and sine functions, like below.

An ellipse centered at {eq}\displaystyle (x_0,y_0) {/eq} and semi-axes a, b, given as {eq}\displaystyle \frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1, {/eq}

is parametrized counterclockwise as {eq}\displaystyle x=x_0+a\cos t, y=y_0+b\sin t, 0\leq t\leq 2\pi, \text{ (for one turn)} {/eq}

or clockwise as {eq}\displaystyle x=x_0+a\sin t, y=y_0+b\cos t, 0\leq t\leq 2\pi {/eq}

A) To find a parametrization for the circle centered at {eq}\displaystyle (2,1) {/eq} and radius 3, counterclockwise,

we will write the equation in Cartesian coordinates, first.

The circle is given in Cartesian coordinates as {eq}\displaystyle (x-2)^{2} + (y-1)^{2} = 9 {/eq}

and to be traced counterclockwise we will choose cosine for the x variable and sine for y,

{eq}\displaystyle x-2=3\cos t, y-1=3\sin t \iff x=2+3\cos t, y=1+3\sin t. {/eq}

So, a parametrization of the circle with the required direction, traced once is {eq}\displaystyle \boxed{x=2+3\cos t, y=1+3\sin t, 0\leq t\leq 2\pi}. {/eq}

B) For three turns, in the same direction, we will use the same form found in A), but we will just extend the interval for t,

therefore a parametrization of the circle traced three times e is {eq}\displaystyle \boxed{x=2+3\cos t, y=1+3\sin t, 0\leq t\leq 6\pi}. {/eq}

C) To find a parametrization for the ellipse {eq}\displaystyle 4(x-5 )^2+9(y+2)^2 = 36 \iff \frac{(x-5 )^2}{3^2}+\frac{(y+2)^2}{2^2} = 1 {/eq}

traced clockwise we will choose sine function for the x variable and cosine for y,

{eq}\displaystyle x-5=3\sin t, y+2=2\cos t \iff x=5+3\sin t, y=-2+2\cos t. {/eq}

So, a parametrization of the ellipse with the clockwise direction, traced once is {eq}\displaystyle \boxed{x=5+3\sin t, y=-2+2\cos t, 0\leq t\leq 2\pi}. {/eq}