How do you graph #r=4+6costheta#?

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Answer 1 · 2017-02-19T16:18:19

See Socratic graph for the Cartesian form of the equation.

This limacon passes through the pole.

The pole is a dimple.

As #r = f(costheta)=f(cos(-theta)#, the graph is symmetrical about

#theta = 0#.

Maximum #r = f(0)=f(2pi)=10#.

I have used the Cartesian form of the polar equation.

#x^2+y^2-4sqrt(x^2+y^2)-6x=0#.

graph{x^2+y^2-4sqrt(x^2+y^2)-6x=0 [-20, 20, -10, 10]}