Question

Question #1b305

Answer 1

`(x, y)=(6, 0)`. If the question is about rectangular form for `r=6 cos theta`, the answer is `x^2+y^2-6x=0`.

Explanation:

Given `(r, theta)=(6, 0)`,

`x = r cos theta=6 cos 0= 6 and y= r sin theta=6 sin 0 =0`.

So, `(x, y)=(6, 0)`

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

`(x-3)^2+y^2=3^2`, revealing center at (3, 0) and radius =3.