# How do you graph the system of polar equations to solve r=2+2costheta and r=3+sintheta?

Feb 5, 2017

See the graph for your pair of a cardioid and a limacon. I have used Socratic utility. Common points are $\left(2 , - \frac{\pi}{2}\right) \mathmr{and} \left(3.79 , {26.57}^{o}\right)$.

#### Explanation:

Here, use cartesian forms of the equations :

For the cardioid, it is from

${r}^{2} = {x}^{2} + {y}^{2} = 2 r + 2 r \cos \theta = 2 \sqrt{{x}^{2} + {y}^{2}} + 2 x$

and, for the limacon, it is from

${r}^{2} = {x}^{2} + {y}^{2} = 3 r + r \sin \theta = 3 \sqrt{{x}^{2} + {y}^{2}} + y$

It is revealed that one common point is on

$\theta = - \frac{\pi}{2}$, at which r = 2.

We can measure the other angle or solve

$r = 2 + 2 \cos \theta = 3 + \sin \theta$. This gives

$\cos \left(\theta + {\cos}^{- 1} \left(\frac{2}{\sqrt{5}}\right)\right) = \frac{1}{\sqrt{5}}$, from which

$\cos \left(\theta + {26.57}^{0}\right) = \cos \left({63.43}^{o}\right)$, and so,

$\theta = {26.57}^{o} ^ \mathmr{and} r = 3.79$, nearly.

For the second point, use

$\theta + {\cos}^{- 1} \left(\frac{2}{\sqrt{5}}\right) = 2 \pi + {\cos}^{- 1} \left(\frac{1}{\sqrt{5}}\right)$, giving

$\theta = {360}^{o} - \left({26.57}^{0} + {63.43}^{o}\right) = {270}^{o}$ that is equivalent to

$- {90}^{o}$.

graph{(x^2+y^2-2sqrt(x^2+y^2)-2x)(x^2+y^2-3sqrt(x^2+y^2)-y)=0 [-10, 10, -5, 5]}