1
$\begingroup$

I have two curves, a circle defined either parametrically or cartesian: $$(x-x_9)^2+(y-y_9)^2-r_9^2=0$$ $$ \left\{ \begin{array}{ll} x = x_9+r_9 \sin(t) \\ y = y_9+r_9 \cos(t) \end{array} \right. $$ and a Limaçon curve parametrically defined by a pedal point $P(x_0,y_0)$ (and another circle): $$ \left\{ \begin{array}{ll} x = x_0\sin^2(s) -\cos(s)\sin(s)(y_0-y_i-r_i\sin(s))+\cos^2(s)(x_i+r_i\cos(s)) \\ y = y_0\cos^2(s) -\cos(s)\sin(s)(x_0-x_i-r_i\cos(s))+\sin^2(s)(y_i+r_i\sin(s)) \end{array} \right. $$

I would like to find the intersection points between the two curves.

Is there any way to do this non-numerically? Is there a way to convert the parametric Limaçon equation to a cartesian form? (it's the normal Limaçon equation, but translated and rotated)

Best regards, Jonas

$\endgroup$

1 Answer 1

0
$\begingroup$

You can plug in the $(x,y)$ Cartesian coordinates of the Limaçon into the equation of the circle. This provides you with a single trigonometric equation with the $s$ variable.

You can rewrite this equation using only the variable $X=\cos s$. Unfortunately, you'll get an equation of degree 6 in $X$. However, you can use any classical method to solve polynomial equations to get the set of $X$ solutions and then $s$.

$\endgroup$
3
  • $\begingroup$ Thanks! I'll try that out. What about $\sin(s)$? How do I relate them to $X$? $\endgroup$
    – lannotje
    Commented Sep 1, 2021 at 10:24
  • $\begingroup$ You have to use usual trigonometric formulas like $\sin^2 s = 1 - \cos^2 s$. $\endgroup$ Commented Sep 1, 2021 at 10:34
  • $\begingroup$ Ah yes thanks! Was just about trying that out :) $\endgroup$
    – lannotje
    Commented Sep 1, 2021 at 11:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .