# How do you convert r=6cos(theta) -8sin(theta) into cartesian form?

Aug 9, 2016

${x}^{2} + {y}^{2} - 6 x + 8 y = 0$.

#### Explanation:

The conversion formula is r(cos theta, sin theta = (x, y)

The given equation is

$r = \frac{6}{r} x - \frac{8}{r} y$. Cross multiplying,

${r}^{2} = {x}^{2} + {y}^{2} = 6 x - 8 y$

So, the cartesian equation is

${x}^{2} + {y}^{2} - 6 x + 8 y = 0$

The standaed form is

${\left(x - 3\right)}^{2} + {\left(y + 4\right)}^{2} = {5}^{2}$.

This represents the circle with center at $\left(3 , - 4\right)$ and radius 5. ,