What the is the polar form of $sqrt(x^2+y^2) = 6x+7y-x^2y-2xy$ ?

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Answer 1 · 2016-10-12T13:50:03

We know the relations

$x=rcostheta and y =rsintheta$

Again $x^2+y^2=r^2$

where r and $theta$ are the polar coordinate of a point having rectangular coordinate $(x,y)$

The given equation in rectanglar form is

$sqrt(x^2+y^2)=6x+7y-x^2y-2xy$

$=>sqrt(r^2)=6rcostheta+7rsintheta-r^3cos^2thetasintheta-2r^2costhetasintheta$

$=>r=6rcostheta+7rsintheta-r^3cos^2thetasintheta-r^2sin2theta$

$=>6costheta+7sintheta-r^2cos^2thetasintheta-rsin2theta=1$

This is the polar form of the given equation.

What the is the polar form of sqrt(x^2+y^2) = 6x+7y-x^2y-2xy sqrt(x^2+y^2) = 6x+7y-x^2y-2xy ?