How do you graph #r=8+6costheta#?

1 Answer
A. S. Adikesavan
Jan 5, 2017

Using Socratic graphic utility, the graph of the limacon could be inserted, for the cartesian form #x^2+y^2=8sqrt(x^2+y^2)+6x#.

Explanation:

#r = 8 + 6 cos theta >=2.#

As# r(theta) = r(-theta)#, the graph is symmetrical about #theta = 0 #.

Range of r is from 2 to 14. 1-period ( #theta in [-pi. pi]#) is inserted.

graph{x^2+y^2-6x-8sqrt(x^2+y^2)=0x^2 [-24, 24, -12, 12]}

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