# parametrizing quarter of a circle

I am given the circle whose equation is: $(x-\frac{1}{2})^{2}+(y+\frac{1}{2})^{2}=\frac{1}{2}$. So, the coordinates of the origin of the circle are: $(\frac{1}{2},-\frac{1}{2})$ and the radius of the circle is : $\frac{1}{\sqrt{2}}$. I am given the points: $A(0,0)$, and $B(1,0)$. Let's consider the parameter $S$ such that when $S=0$, we are at the point $A$, and when $S=1$, we are at the point $B$. I need to parametrize the quarter of the circle that starts from A and ends at B, in terms of the parameter $S$ (i.e, I am trying to find $x(S)$ and $y(S)$). I tried many times, but I couldn't come up with expresions for $x(S)$ and $y(S)$ that satisfy the equation of the circle, and such that when $S=0$, we have : ($x=0$ and $y=0$), and when $S=1$, we have $x=1$ and $y=0$. Any help is appreciated!!!

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You can obtain such a parametrization by starting with something you know and then transforming it to work with your circle.

More concretely, say the circle is the unit circle centered at origin, and say we're going from $A = (1,0)$ to $B = (0,1)$, which sweeps $90^\circ$ clockwise. The parametrization for this would be

$x(S) = \cos(\frac{\pi}{2} S)$

$y(S) = \sin(\frac{\pi}{2} S)$

But in your circle, the parametrization is supposed to sweep counterclockwise, so we can parametrize backwards by replacing $S$ with $1-S$:

$x(S) = \cos(\frac{\pi}{2} (1-S))$

$y(S) = \sin(\frac{\pi}{2} (1-S))$

But this sweeps from $\theta = \frac{\pi}{2}$ to $\theta = 0$, whereas you want $\theta = \frac{3\pi}{4}$ to $\theta = \frac{\pi}{4}$. To adjust for this, we note that this is simply our circle rotated clockwise by $\frac{\pi}{4}$, so add $\frac{\pi}{4}$ to the arguments of sine and cosine in the parametrization:

$x(S) = \cos(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$

$y(S) = \sin(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$

Now all we're missing is the new radius and center of the circle. To adjust for the radius, just scale $x$ and $y$ by the radius:

$x(S) = \frac{1}{\sqrt 2} \cos(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$

$y(S) = \frac{1}{\sqrt 2} \sin(\frac{\pi}{2}(1-S) + \frac{\pi}{4})$

Finally, to translate the circle so that its center is at $(\frac12, -\frac12)$ instead of $(0,0)$, we must add $\frac12$ to $x$ and $-\frac12$ to $y$, so we end up with

$x(S) = \frac{1}{\sqrt 2} \cos(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) + \frac12$

$y(S) = \frac{1}{\sqrt 2} \sin(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) - \frac12$.

You can check that this indeed gives you want you want. The starting point is

$(x(0),y(0)) = (\frac{1}{\sqrt 2} \cos(\frac{3\pi}{4}) + \frac12, \frac{1}{\sqrt 2} \sin(\frac{3\pi}{4}) - \frac12) = (0,0)$

and the endpoint is

$(x(1),y(1)) = (\frac{1}{\sqrt 2} \cos(\frac{\pi}{4}) + \frac12, \frac{1}{\sqrt 2} \sin(\frac{\pi}{4}) - \frac12) = (1,0)$

and moreover our parametrized curve stays on the circle since

$(x(S) - \frac12)^2 + (y(S) + \frac12)^2 = \frac12 \cos^2(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) + \frac12 \sin^2(\frac{\pi}{2}(1-S) + \frac{\pi}{4}) = \frac12$

for all $S \in [0,1]$.

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Thanks a lot for this beautiful and detailed proof – C. Lambda Nov 4 '12 at 18:12