Polar form to cartesian

Question

Let $\Gamma$ be a circle that passes through the origin. Show that we can find real numbers $s$ and $t$ such that $\Gamma$ is the graph of $r = 2s \cos (\theta + t).$

I know this has to be converted to a cartesian equation, but how do I do this, and what do I do after?

Thanks

Answer 1

You have $x(\theta)=r(\theta) \cos \theta$ and $y(\theta)=r(\theta) \sin \theta$. Based on that you can use trigonometric formulaes to compute the cartesian coordinates $(x(\theta),y(\theta))$ and recognize the equation of a circle.

Answer 2

A circle that passes through the origin can be expressed as $$(x-a)^2+(y-b)^2=a^2+b^2,$$ i.e. $$x^2-2ax+y^2-2by=0$$ for some $a,b\in\mathbb R$ such that $(a,b)\not=(0,0)$.

Here, setting $x=r\cos\theta,y=r\sin\theta$ gives you $$r^2\cos^2\theta-2ar\cos\theta+r^2\sin^2\theta-2br\sin\theta=0,$$ i.e. $$r=2a\cos\theta+2b\sin\theta.$$ Since this can be written as $$r=2\sqrt{a^2+b^2}\cdot \cos\theta\cdot \frac{a}{\sqrt{a^2+b^2}}-2\sqrt{a^2+b^2}\cdot\sin\theta\cdot\left(-\frac{b}{\sqrt{a^2+b^2}}\right)$$ setting $s=\sqrt{a^2+b^2},\cos t=\frac{a}{\sqrt{a^2+b^2}},\sin t=-\frac{b}{\sqrt{a^2+b^2}}$ gives you $$r=2s\cos(\theta+t).$$