How do you convert #r=6cos(theta) -8sin(theta)# into cartesian form?

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Answer 1 · 2016-08-09T16:02:45

#x^2+y^2-6x+8y=0#.

The conversion formula is #r(cos theta, sin theta = (x, y)#

The given equation is

#r = 6/r x-8/r y#. Cross multiplying,

#r^2 = x^2 + y^2 = 6 x - 8 y#

So, the cartesian equation is

#x^2+y^2-6x+8y=0#

The standaed form is

#(x-3)^2+(y+4)^2=5^2#.

This represents the circle with center at #(3, -4)# and radius 5. ,