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Eliminate the parameter. x = 3 cos t, y = 3 sin t
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First, square each equation: \[x^2=9\cos^2 t\]\[y^2=9\sin^2 t\]Adding the two equations, we have \[x^2+y^2=9\cos^2 t+9\sin^2 t=9(\cos^2 t+\sin^2 t)\]Is the expression inside the parentheses familiar to you? This is automatically equal to 1, and thus the result will be \(x^2+y^2=9\).
Yea, it's familiar.
Thanks! :) it helps alot!
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