# How can I convert $r = 2\cos{\theta}+2\sqrt{3}\sin{\theta}$ to cartesian coordinates? [closed]

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?

## closed as off-topic by Eevee Trainer, verret, воитель, Leucippus, ShaileshJun 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, $$(0, 2b), (2a, 0), (2a, 2b)$$.