How do you convert #(x + 1)^2 + (y + 1)^2 =4# into polar form?

Substitute #x = r cos theta# and #y = r sin theta# into the equation to get:

#(r cos theta + 1)^2 + (r sin theta + 1)^2 = 4#

Hence:

#r^2 + 2(cos theta + sin theta)r - 2 = 0#

Explanation:

Substitute #x = r cos theta# and #y = r sin theta# into the equation to get:

#(r cos theta + 1)^2 + (r sin theta + 1)^2 = 4#

This can be reformulated as follows:

#4 = (r cos theta + 1)^2 + (r sin theta + 1)^2#

#= r^2 cos^2 theta + 2r cos theta + 1 + r^2 sin^2 theta + 2r sin theta + 1#

#= r^2 (cos^2 theta + sin^2 theta) + 2r (cos theta + sin theta) + 2#

#= r^2 + 2r (cos theta + sin theta) + 2#

Subtract #4# from both ends to get:

#r^2 + 2(cos theta + sin theta)r - 2 = 0#