How do you convert this polar equation to cartesian form?

Question

#r=36/5sin(theta)+36cos(theta)#
convert into the rectangular form

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Answer 1 · 2018-04-18T19:27:50

Given: #r=36/5sin(theta)+36cos(theta)#

Multiply both sides by #r#:

#r^2=36/5rsin(theta)+36rcos(theta)#

Substitute #r^2= x^2+y^2#, #rsin(theta) =y#, and #rcos(theta) = x#:

#x^2+y^2=36/5y+36x#

Rewrite equal to 0:

#x^2-36x+y^2-36/5y= 0#

Add #h^2+k^2# to both sides:

#x^2-36x+h^2+y^2-36/5y+k^2= h^2+k^2#

From the pattern #(x-h)^2=x^2-2hx+h^2# we observe that:

#-2hx = -36x#

#h = 18#

#(x-18)^2+y^2-36/5y+k^2= 18^2+k^2#

From the pattern #(y-k)^2= y^2-2ky+k^2# we observe that:

#-2ky = -35/5y#

#k = 35/10#

#(x-18)^2+(y-36/10)^2= 18^2+(36/10)^2#

Write the right side as a square:

#(x-18)^2+(y-36/10)^2= (18/5sqrt26)^2#

This is a circle of radius, #18/5sqrt26#, with its center at #(18,36/10)#