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

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Answer 1 · 2016-10-26T20:12:57

Expand the squares:

#4 = x^2 + 6x + 9 + 4y^2 - 16y + 16#

Combine the constants and group the square terms together:

#x^2 + 4y^2 + 6x - 16y + 21 = 0#

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

#(cos^2(theta) + 4sin^2(theta))r^2 + (6cos(theta) - 16sin(theta))r + 16 = 0#

This is a quadratic in r, use the quadratic formula:

#r = {(16sin(theta) - 6cos(theta)) +-sqrt(((6cos(theta) - 16sin(theta)))^2 - 64(cos^2(theta) + 4sin^2(theta)))}/(2(cos^2(theta) + 4sin^2(theta)))#