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. ,

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