<s>[INST] Derive that $r=a\cos θ$ from $(x-h)^2 + (y-k)^2 = r^2$ I am a beginner here so please kindly tell me how I can do better when asking my questions so I can improve next time and if there are any complaints to this question.
I want to ask, how do I obtain the formula $r=a\cos θ$ from the circle formula $(x-h)^2 + (y-k)^2 = r^2$.
More preferably how do I obtain it algebraically?
This problem I have comes from I believe Polar Curves/Equations; where the circle is tangent to the $\frac{1}{2}π$ axis with diameter $a$.
Please explain to me how you would have taught it to yourself if you were new to this kind of problem.
 [/INST] $$x=r \cos\theta\space; y=r \sin\theta\space; r=\frac{a}{2}\space;
C(\frac{a}{2}, 0)$$
$$ (x-h)^2 + (y-k)^2 = r^2$$
$$ (r \cos\theta-\frac{a}{2})^2 + (r \sin\theta-0)^2=(\frac{a}{2})^2$$
Using the identity $ (a-b)^2 = a^2-2ab+b^2$
$$
r^2\cos^2\theta - a r \cos\theta + \frac{a^2}{4} + r^2\sin^2\theta
=\frac{a^2}{4}
$$
Simplifying and rearranging the terms we get...
$$r^2\cos^2\theta + r^2\sin^2\theta - a r \cos\theta = 0$$
$$r^2(\cos^2\theta+\sin^2\theta) - a r \cos\theta = 0$$
$ \cos^2\theta + \sin^2\theta = 1$
$$r^2 - a r \cos\theta = 0$$
$$ r^2 = a r \cos\theta$$
$$ r = a \cos\theta$$
Tried to explain it the way I understood from all the responses, thanks guys!
</s>