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?
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What other circle equation do you know ? And are you familiar with the trigonometric circle ? – Aug 09 '17 at 14:39
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Can you give more detail, either on what you are looking for or on the problem? From the information given it looks like you have the answer already ($x=a-a\cos(\theta),\ y=a-a\sin(\theta)$). – Cole Aug 09 '17 at 14:41
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I am familiar with the trigonometric circle and polar coordinates. I know $ x^2+y^2 = r^2$ where r is the radius and for a circle at coordinate (h,k) its $ (x-h)^2+(y-k)^2 = r^2$. But I don't know how they find $ ( x=a−acos(\theta), y=a−asin(\theta)$ – Ayoub Rossi Aug 09 '17 at 14:44
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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 $y$.
marty cohen
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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 $\theta+\pi$,
$$(x,y)=(a+a\cos(\theta+\pi),a+a\sin(\theta+\pi))=(a-a\cos(\theta),a-a\sin(\theta)).$$
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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 $$0\le u\le \frac {\pi}{2} $$
hamam_Abdallah
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x = a+acos(t) or x = a-acos(t)? and y = a+asin(t) or y = a-asin(t) – Ayoub Rossi Aug 09 '17 at 15:07
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