How do you convert #1=(x+4)^2+(y+7)^2# into polar form?

1 Answer
Bio
Apr 2, 2016

#r = +-sqrt{(4 cos(theta) + 7 sin(theta))^2-64} - (4 cos(theta) + 7 sin(theta))#

Explanation:

Let #x = r cos(theta)# and #y = r sin(theta)#.

#1 = (x + 4)^2 + (y + 7)^2#

#1 = (r cos(theta) + 4)^2 + (r sin(theta) + 7)^2#

#1 = (r^2 cos^2(theta) + 8 r cos(theta) + 16) #

#+ (r^2 sin^2(theta) + 14 r sin(theta) + 49)#

#1 = r^2 (cos^2(theta) + sin^2(theta)) + r (8 cos(theta) + 14 sin(theta)) + 65 #

#0 = r^2 + 2r (4 cos(theta) + 7 sin(theta)) + 64 #

Complete the square

#(r + (4 cos(theta) + 7 sin(theta)))^2 + 64 = (4 cos(theta) + 7 sin(theta))^2#

#r = +-sqrt{(4 cos(theta) + 7 sin(theta))^2-64} - (4 cos(theta) + 7 sin(theta))#

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