Mathematics - Circle Question with Solution | TestHub

Given circle$x^2+y^2-5x-y+5=0$
$\Rightarrow {\left(x-\frac{5}{2}\right)}^2+{\left(y-\frac{1}{2}\right)}^2=\frac{3}{2}$
On circle$Q=(\frac{5}{2}+\sqrt{\frac{3}{2}}\cos \theta ,\frac{1}{2}+\sqrt{\frac{3}{2}}\sin \theta )$

$P{Q}^2={\left(\frac{5}{2}+\sqrt{\frac{3}{2}}\cos \mathrm{\theta }\right)}^2+{\left(\frac{5}{2}+\sqrt{\frac{3}{2}}\sin \mathrm{\theta }\right)}^2$

$P{Q}^2=\frac{25}{2}+\frac{3}{2}+5\sqrt{\frac{3}{2}}\left(\cos \mathrm{\theta }+\sin \mathrm{\theta }\right)$

$=14+5\sqrt{\frac{3}{2}}\left(\cos \mathrm{\theta }+\sin \mathrm{\theta }\right)$

$\therefore$Maximum value of${\left(PQ\right)}^2=14+5\sqrt{\frac{3}{2}}\left(\sqrt{2}\right)$

$=14+5\sqrt{3}$.