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I'm struggling to figure out this circle equation in polar coordinates:

$r = 2\cos{\theta}+2\sqrt{3}\sin{\theta}$

and converting it to cartesian form.

How can I convert this to cartesian? How can I tell its radius and center point in both polar and cartesian forms?

Thanks in advance!

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closed as off-topic by Eevee Trainer, verret, воитель, Leucippus, Shailesh Jun 21 at 2:27

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Just apply $x = r\cos(\theta), y=r\sin(\theta) $ and $r^2 = x^2+y^2$.

$r = 2\cos(\theta)+2\sqrt{3}\sin(\theta) $ and $r^2 = x^2+y^2 $ so $r^2 = 2r\cos(\theta)+2\sqrt{3}r\sin(\theta) =2x+2\sqrt{3}y =x^2+y^2 $ so $x^2-2x+y^2-2\sqrt{3}y =0 $ or $x^2-2x+1+y^2-2\sqrt{3}y+3 =4 $ or $(x-1)^1+(y-\sqrt{3})^2 =4 $.

More generally, if $r = 2a\cos(\theta)+2b\sin(\theta) $, then $r^2 = 2ar\cos(\theta)+2br\sin(\theta) =2ax+2by =x^2+y^2 $, so $(x-a)^2+(y-b)^2 =a^2+b^2 $.

This is a circle with center $(a, b)$ and radius $\sqrt{a^2+b^2}$. It passes through the origin, $(0, 2b), (2a, 0), (2a, 2b)$.

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  • $\begingroup$ Sorry @marty-cohen, there was a typo, the equation has a plus between right hand terms. Corrected now. $\endgroup$ – Regis Fontes Jun 20 at 22:15
  • $\begingroup$ OK. Made it simpler. $\endgroup$ – marty cohen Jun 20 at 22:19
  • $\begingroup$ Thank you so so much. Specially for making it general! $\endgroup$ – Regis Fontes Jun 20 at 22:25
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    $\begingroup$ I always prefer to solve an infinite number of problems rather than just one. $\endgroup$ – marty cohen Jun 20 at 22:26

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