# Question 1b305

May 10, 2016

$\left(x , y\right) = \left(6 , 0\right)$. If the question is about rectangular form for $r = 6 \cos \theta$, the answer is ${x}^{2} + {y}^{2} - 6 x = 0$.

#### Explanation:

Given $\left(r , \theta\right) = \left(6 , 0\right)$,

$x = r \cos \theta = 6 \cos 0 = 6 \mathmr{and} y = r \sin \theta = 6 \sin 0 = 0$.

So, $\left(x , y\right) = \left(6 , 0\right)$

Thanks to Sente, for pointing out the following aspect of the question, I am now adding this part now.

If the question is about rectangular form for $r = 6 \cos \theta$m, use r^2=x^2+y^2.

When theta=0, r=0 and so are x and y.

Elsewhere,)

r=6x/r. So, r^2=x^2+y^2=6. and the answer is x^2+y^2-6x=0.

Center-radius form of this equation of the circle through the origin is

${\left(x - 3\right)}^{2} + {y}^{2} = {3}^{2}$, revealing center at (3, 0) and radius =3.