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:

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

How do you convert 4=(-x-3)^2+(2y-4)^24=(-x-3)^2+(2y-4)^2 into polar form?