# How do you convert 9r=4cos(theta) into cartesian form?

$9 {x}^{2} + 9 {y}^{2} - 4 x = 0.$
To convert from Polar form to Cartesian, we need the following formula $\left(1\right) \cos \theta = \frac{x}{r} , \left(2\right) \sin \theta = \frac{y}{r} , \left(3\right) r = \sqrt{{x}^{2} + {y}^{2}} .$
Using (1) in the given polar eqn. we have$9 r = 4 \frac{x}{r} , \mathmr{and} , 9 {r}^{2} = 4 x ,$ & now by (3), $9 \left({x}^{2} + {y}^{2}\right) = 4 x , i . e . , 9 {x}^{2} + 9 {y}^{2} - 4 x = 0.$