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Question #69a2e

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Dec 14, 2016

Given $r^{2} = θ$ $r^2 = theta$

The breaks into 3 equations depending on the value of x and y:

$x^{2} + y^{2} = {tan}^{- 1} (\frac{y}{x}); x > 0, y > 0$ $x^2 + y^2 = tan^-1(y/x); x > 0, y > 0$

$x^{2} + y^{2} = π + {tan}^{- 1} (\frac{y}{x}); x < 0$ $x^2 + y^2 = pi + tan^-1(y/x); x < 0$

$x^{2} + y^{2} = 2 π + {tan}^{- 1} (\frac{y}{x}); x > 0, y < 0$ $x^2 + y^2 = 2pi + tan^-1(y/x); x > 0, y < 0$

$x = 0$ $x = 0$ when you choose $y = 0$ $y = 0$ but not the vice-versa.

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