How can you determine that the polar equation $r = a\cos(\theta)$ is a circle?
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2 Answers
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$r^2=ar\cos\theta$. Now use the substitutions $r^2=x^2+y^2$ and $x=r\cos\theta$.
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Consider $$ \begin{align} y^2+(x-a/2)^2 &=\color{#C00000}{(r\sin(\theta))^2}+\color{#00A000}{(r\cos(\theta)-a/2)^2}\\ &=\color{#C00000}{a^2\cos^2(\theta)\sin^2(\theta)}+\color{#00A000}{a^2\cos^4(\theta)-a^2\cos^2(\theta)+a^2/4}\\ &=\color{#0000FF}{a^2\cos^2(\theta)(\sin^2(\theta)+\cos^2(\theta))}-a^2\cos^2(\theta)+a^2/4\\ &=\color{#0000FF}{a^2\cos^2(\theta)}-a^2\cos^2(\theta)+a^2/4\\ &=a^2/4 \end{align} $$
