Polar Equation to Cartesian Equation.

Find the cartesian equation of the circle with polar equation $r=2a\cos \theta$

My attempt,

Since $\cos \theta=\frac{x}{r}$

So, $r=2a(\frac{x}{r})$

I don't know how to proceed.

4 Answers

HINT: we have $$r=\sqrt{x^2+y^2}$$

Are you sure it is a circle? I would expect the equation of a circle to be $r = a$. Recall that $r^2 = x^2 + y^2$. \begin{eqnarray*} \sqrt { x^2 + y^2 } &=& 2a \frac{x}{r} \\ x^2 + y^2 &=& 4a^2 \frac{x^2}{r^2} \\ x^2 - 4a^2 \frac{x^2}{r^2} + y^2 &=& 0 \\ (1 - \frac{4a^2}{r^2})x^2 + y^2 &=& 0 \\ \end{eqnarray*}

$r = 2a(x/r)$ implies $r^2 = 2ax$ but $r^2 = x^2 + y^2$ as well. Combining these gives $x^2 - 2ax + y^2 = 0$ and tidying up gives $(x-a)^2 + y^2 = a^2$

You know that $$x=r\cos\theta,\quad y=r\sin\theta$$ We can substitute in for $r$: $$x=2a\cos^2\theta,\quad y=2a\cos\theta\sin\theta$$ We can say that $$\cos\theta=\sqrt{\frac{x}{2a}}$$ and, by the identity $\sin^2\theta+\cos^2\theta=1$, $$\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-\frac{x}{2a}}$$ You now have expressions for $\cos\theta$ and $\sin\theta$ as functions of $x$, which you can plug into the equation for $y$: $$y=2a\sqrt{\frac{x}{2a}}\sqrt{1-\frac{x}{2a}}$$ All you have to do now is simplify. You can multiply the roots together, then try to change the expression inside the square root to resemble a more familiar equation.