Question #1b305

1 Answer
May 10, 2016

Answer:

#(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.