Question

Lets say I have a circle centred at X=-1 The radius is 1, how can I express r? The area integral will need to be expressed in terms of drdt where t ranges from Pi/2 to 3Pi/2

Answer 1

Unit circle t: 0 to 2Pi but in the other case when r: 0 to -2 t:Pi/2 to 3Pi/2

Answer 2

\[\int\limits \!r\,{\rm d}[r={0\ldots -2\,\cos \left( t \right) },t={\pi/2 \ldots 3/2\,\pi}]\] I know excplicit for this circle, but more in general, how can i express r if the circle isnt centred at orgin

Answer 3

Consult the standard form of the circle equation:
\[(x-a)^2+(y-b)^2=r^2\]is a circle centered at \((a,b)\) with radius \(r\). Converting to polar coordinates with radius \(1\) yields
\[\begin{align*}
(x+1)^2+y^2=1\implies (r\cos\theta+1)^2+(r\sin\theta)&=1\\[1ex]
r^2\cos^2\theta+2r\cos\theta+1+r^2\sin^2\theta&=1\\[1ex]
r^2+2r\cos\theta&=0
\end{align*}\]We require \(r>0\), so we can divide through by \(r\) to get
\[r+2\cos\theta=0\implies r=-2\cos\theta\]

Answer 4

And in general, with arbitrary radius \(R\) and center \((a,b)\), you end up with
\[\begin{align*}
R^2&=(r\cos\theta-a)^2+(r\sin\theta-b)^2\\[1ex]
R^2&=r^2\cos^2\theta-2ar\cos\theta+a^2+r^2\sin^2\theta-2br\sin\theta+b^2\\[1ex]
R^2-a^2-b^2&=r^2-2r(a\cos\theta+b\sin\theta)\\[1ex]
R^2-a^2-b^2&=\bigg(r-(a\cos\theta+b\sin\theta)\bigg)^2-(a\cos\theta+b\sin\theta)^2
\end{align*}\]and so on.

Answer 5

Amaizing, you never fail to impress :)