How do you convert #(x-3)^2+(y+4)^2=25# into polar form?

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Answer 1 · 2016-11-19T02:05:42

Please see the explanation for steps leading to:

#r = 6cos(theta) - 8sin(theta)#

Expand the squares:

#x^2 - 6x + 9 + y^2 + 8y + 16 = 25#

Group the square terms, the linear terms and please notice that the constants sum to zero:

#x^2 + y^2 - 6x + 8y = 0#

Substitute #r^2# for #x^2 + y^2#

#r^2 - 6x + 8y = 0#

Substitute #rcos(theta)# for x and #rsin(theta)# for y:

#r^2 - 6rcos(theta) + 8rsin(theta) = 0#

Discard a common r factor:

#r - 6cos(theta) + 8sin(theta) = 0#

Move the sine and cosine terms to the right side:

#r = 6cos(theta) - 8sin(theta)#