How do you find the rectangular equation for #r=-6sintheta#?

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Answer 1 · 2016-11-12T17:09:30

Please see the explanation for steps leading to the equation of a circle:

#(x - 0)^2 + (y - -3)^2 = 3^2#

Multiply both sides of the equation by r:

#r^2 = -6rsin(theta)#

Substitute #x^2 + y^2# for #r^2# and y for #rsin(theta)#:

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

Add #6y + k^2# to both sides:

#x^2 + y^2 + 6y + k^2 = k^2#

Use the right side of the pattern #(y - k)^2 = y^2 - 2ky + k^2#, to complete the square for the y terms:

#y^2 - 2ky + k^2 = y^2 + 6y + k^2#

#-2ky = 6y#

#k = -3# and #k^2 = 9 = 3^2#

Replace the y terms with the left side of the pattern but with #k = -3#:

#x^2 + (y - -3)^2 = k^2#

To put this into the standard form for a circle, substute #3^2# for #k^2# (not 9) )and insert a #-0# in the x term:

#(x - 0)^2 + (y - -3)^2 = 3^2#

This is a circle with its center at #(0, -3)# and a radius of 3.