Answer
$c^{2}=a^{2}+b^{2}-2ab cos \theta$
Work Step by Step
Here $x=bcos\theta$ and $y=bsin\theta$
Thus $P(x,y)=P(bcos\theta,bsin\theta)$
Use the distance formula to compute $c$.
Therefore,
$c=\sqrt {(a-bcos\theta)^{2}+(0-bsin \theta)^{2}}$
$=\sqrt {(a^{2}+b^{2}cos^{2}\theta-2ab cos \theta+b^{2}sin ^{2}\theta}$
$=\sqrt {(a^{2}-2ab cos \theta+b^{2}(1)}$
Hence, $c^{2}=a^{2}+b^{2}-2ab cos \theta$