How to find the parametric equation

Question

I've got this curve and I know its part of a circle,

From this I should find that $x = a-acos(\theta) $and $y = a-asin(\theta)$ but I don't know how?

Can someone help?

Answer 1

The quarter-circle in the third quadrant centered at the origin is $x=-a\cos t, y=-a\sin t$.

Your curve is that with $a$ added to $x$ and $y$.

Answer 2

From the figure, the radius is $a$ and the center is at $(a,a)$.

To the center coordinates, add the components of the radius at angle $\theta+\pi$,

$$(x,y)=(a+a\cos(\theta+\pi),a+a\sin(\theta+\pi))=(a-a\cos(\theta),a-a\sin(\theta)).$$

Answer 3

The circle has the equation :

$$(x-a)^2+(y-a)^2=a^2$$

so the parametric equations of the portion are

$$x=a+a\cos (t) $$ $$y=a+a\sin (t) $$ with $$\pi \le t \le \frac {3\pi}{2} $$

or $$x=a-a\cos (u) $$ $$y=a-a\sin (u) $$ with $$0\le u\le \frac {\pi}{2} $$