Find the parametric equation of a circumference in a non traditional reference frame

Question

I know that the parametric equation of a circumference center in (0,0) is:

$y=\sin(\alpha)*r, x = \cos(\alpha)*r$.

I try to write the equation of the same circumference with respect to the reference frame in the figure ($x',y'$) -- center of the frame $y_C=-0.6, \,x_C=0$.

I try in a lot of ways but I can't find an answer.

How can I write it?

I think that it should be something like:

$x=f_1(\theta,r)*\cos(\theta), y=f_2(\theta,r)*\sin(\theta)$

Answer 1

In the Cartesian system $(x',y')$, the equation of the circle is $$x'=r\cos\alpha\\y'=0.6+r\sin\alpha$$ From here $$x'^2+(y'-0.6)^2=r^2$$ Using $$\tan\theta=\frac{y'}{x'}$$ We can write $$y'=x'\tan\theta$$ Therefore $$x'^2+(x'\tan\theta-0.6)^2=r^2$$ This is a quadratic. Choose the positive solution at $\theta=0$ $$x'^2(1+\tan^2\theta)-1.2x'\tan\theta+0.36-r^2=0\\ 1+\tan^2\theta=\frac{1}{\cos^2\theta}\\ x'_{1,2}=\frac{1.2\tan\theta\pm\sqrt{1.44\tan^2\theta-4\frac1{\cos^2\theta}(0.36-r^2)}}{\frac2{\cos^2\theta}}$$ You can simplify this further ( I just choose the solution with +): $$x'=\cos\theta\frac{\cos\theta}2\left(1.2\frac{\sin\theta}{\cos\theta}+\frac{2}{\cos\theta}\sqrt{\frac{\cos^2\theta}4(1.44\tan^2\theta+\frac{4}{\cos^2\theta}(r^2-0.36))}\right)\\=\cos\theta(0.6\sin\theta+\sqrt{r^2-0.36\cos^2\theta})$$ Similarly, you can find out $y'$. You should check if I did not make some silly mistakes.

Answer 2

Note that in $x'y'$ this is $$(x')^2+(y'-0.6)^2=r^2$$ Then $x'=r\cos\theta$, and $y'=0 .6+r\sin\theta$

Answer 3

From the sketch (allow me to change your symbology)

it is clear that $$ \left\{ \matrix{ R\cos \theta = r\cos \alpha \hfill \cr R\sin \theta = s + r\sin \alpha \hfill \cr} \right. $$

Squaring each line , summing, and replacing $\alpha$ with $\theta$ $$ \eqalign{ & R^{2} = r^{2} \cos ^{2} \alpha + \left( {s + r\sin \alpha } \right)^{2} = \cr & = r^{2} + s^{2} + 2sr\sin \alpha = r^{2} + s^{2} + 2s\left( {R\sin \theta - s} \right)\quad \Rightarrow \cr & \Rightarrow \quad R^{2} - 2s\sin \theta R - \left( {r^{2} - s^{2} } \right) = 0 \cr & R = s\sin \theta \pm \sqrt {s^{2} \sin ^{2} \theta + r^{2} - s^{2} } \quad \Rightarrow \cr & \Rightarrow \quad R = s\sin \theta + \sqrt {r^{2} - s^{2} \cos ^{2} \theta } \cr} $$

and the last is the polar equation of the excentric circle.