Question

How do you find a parametric equation for a particle moving twice counter-clockwise around the circle `(x-2)^2 + (y+1)^2 = 9` starting at (-1,-1)?

Answer 1

`{: (x = 2+ 3cost),(y = -1 + 3sint) :} } \ \ \pi le t le 5pi`

Explanation:

The Cartesian equation of a circle with centre `(a,b)` and radius `r` is:

`(x-a)^2 + (y-a)^2 = r^2`

And so the equation:

`(x-2)^2 + (y+1)^2 = 9`

represents a circle of centre `(2,-1)` and radius `3`

The parametric equations of a circle of centre `(0,0)` and radius `r` are;

`x = rcost`
`y = rsint`

This can easily be verified, as:

`x^2 + y^2 = (rcost)^2 + (rsint)^2`
`" "= r^2cos^2t + r^2sin^2t`
`" "= r^2(cos^2t + sin^2t)`
`" "= r^2 \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (because cos^2t + sin^2t -= 1)`

And so the parametric equations of circle centre `(a,b)` and radius `r` are simply:

`x = a+ rcost`
`y = b+ rsint`

And so the required equations are:

`x = 2+ 3cost`
`y = -1 + 3sint`

(NB If we swap the `sin"/"cos` we can change the direction of travel; as it happens the order we have chosen will cause an anti-clockwise trajectory).

We require that particle to start at `(-1,-1)` which correspond to a start parameter `t=pi` relative to the centre `(2,-1)`, and we need two full revolutions, and so the end parameter is `t=pi+2*2pi=5pi`

Hence we have:

`{: (x = 2+ 3cost),(y = -1 + 3sint) :} } \ \ \pi le t le 5pi`

The following shows the plot
https://www.desmos.com/calculator/a15yobsjcj