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i have to eliminate the parameter and obtain the standard form of the rectangular equation of x=h+rcos(theta), y=k+rsin(theta)
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$$ x=h+rcos(\theta) \implies cos(\theta)=\cfrac{{(x-h)}^2}{r^2}\\ y=k+rsin(\theta) \implies sin(\theta)=\cfrac{{(y-k)}^2}{r^2}\\ \text{now keeping in mind the pythagorean identity } cos(\theta)^2+sin(\theta)^2=1\\ \cfrac{(x-h)^2}{r^2}+\cfrac{(y-k)^2}{r^2}=1 $$
$$ x=h+rcos(\theta) \implies implies cos(\theta)=\cfrac{{(x-h)}}{r}\\ y=k+rsin(\theta) \implies sin(\theta)=\cfrac{{(y-k)}}{r}\\ \text{now keeping in mind the pythagorean identity } cos(\theta)^2+sin(\theta)^2=1\\ \cfrac{(x-h)^2}{r^2}+\cfrac{(y-k)^2}{r^2}=1 $$ there, better :)
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