How do you write the given equation #(x-3)^2+y^2=9# into polar form?

3 Answers
Jul 18, 2017

#r=6cos theta#

Explanation:

Rewrite the given equation as #x^2 -6x +9 +y^2 =9#. Put #x=r cos theta and y= r sin theta#.

This would make #x^2 +y^2 =r^2# and the equation is thus changed to #r^2 -6rcos theta =0# , Or # r(r-6 cos theta)=0#

Since #r !=0# , the required equation would be #r=6 cos theta#

Jul 18, 2017

You substitude #x=rcosθ# and #y=rsinθ#

Explanation:

#(x-3)^2+y^2=9 => (rcosθ-3)^2+(rsinθ)^2=9=>#

#r^2cos^2θ-6rcosθ+9+r^2sin^2θ=9 => r^2-6rcosθ=0 =>#

#r(r-6cosθ)=0=>r-6cosθ=0 =>r=6cosθ#

Jul 18, 2017

Expand the square.
Use the conversion equations #x^2+y^2=r^2 and x = rcos(theta)#
Express r as a function of #theta#

Explanation:

Given: #(x-3)^2+y^2=9#

Here is the graph of the original equation:

www.Desmos.com/calculator

Expand the square:

#x^2-6x+9+y^2=9#

Use the conversion equations #x^2+y^2=r^2 and x = rcos(theta)#:

#r^2-6rcos(theta)+9=9#

Begin the process of expressing #r# as a function of #theta#

Combine like terms:

#r^2-6rcos(theta)=0#

We can safely divide both sides by r, because we are only eliminating the trivial root #r = 0#:

#r-6cos(theta)=0#

Add #6cos(theta)# to both sides and we shall have #r# as a function of #theta#:

#color(blue)(r=6cos(theta)) larr# The Polar equivalent.

Here is the graph in polar coordinates:

www.Desmos.com/calculator

Please observe that the graphs are identical.