To convert rectangular to polar **in general:**

Plot the point #(x,y)# on the coordinate system.

Draw the line segment fro the origin to #(x,y)#.

Call the length of that line segment #r#. (Note that #r# and is also the distance between the point #(x,y)# and the origin.)

Use the Pythagorean Theorem (or its offspring, the distance formula) to see that:

#r = sqrt(x^2+y^2)#

Find the smallest positive angle from the positive #x# axis to the line segment you drew. Call that angle #theta#

Now from one of the definitions of the trigonometric functions:

If the point #(x,y)# lies on the terminal side of #theta# and #r = sqrt(x^2+y^2)#, then

#sin theta = y/r" "# this gives us the conversion: #y = rsintheta#

#cos theta = x/r" "# this gives us the conversion: #x = rcostheta#

#tan theta = y/x" "# this gives us the conversion to find #theta#

**In this question** , we have:

#x^2+y^2 = 36#

**The long way** is to replace #x# and #y# by #rcos theta# and #rsin theta# and simplify:

#x^2+y^2 = 36#

#(rcos theta)^2+(rsin theta)^2 = 36#

#r^2cos^2 theta+r^2sin^2 theta = 36#

#r^2(cos^2 theta+sin^2 theta) = 36#

#r^2(1) = 36#

#r^2 = 36#

With the usual definition #r# is non-negative, so we get:

#r=6#

**The short way** "Hey, look! The left is just the same as #r^2#"

We immediately get: #r^2 = 36# so #r = 6#.

**The geometric way**

The equation #x^2 +y^2 = 36# has a graph that is the circle, centered at the origin with radius #6#.

So in terms of the polar coordinates, #theta# can be anything it wants to be and #r# must always be #6#.

The equation is #r = 6#