If $p$ and $q$  be the longest and the shortest distances respectively of the point $(-7, 2)$) from any point $(\alpha ,\beta )$  on the circle $x^2+y^2-10x-14y-51=0$. If S is the geometric mean of $p$ and $q$  then $\frac{s}{\sqrt{11}}$ is equal to

Centre of the circle is C(5, 7). Let P(-7, 2) be the given point and the line joining P and C  meet the circle at A and B
 
 Now  $PC=\sqrt{{\left(12\right)}^2+{\left(5\right)}^2}=13$
 Radius of the circle is  $\sqrt{5^2+7^2+51}=5\sqrt{5}$
 $p=PB=13+5\sqrt{5},q=PA=13-5\sqrt{5}$    
   G.M. of  $p,q=\sqrt{pq}=\sqrt{{\left(13\right)}^2-{\left(5\sqrt{5}\right)}^2}=2\sqrt{11}=s$
   $\Rightarrow \frac{s}{\sqrt{11}}=2$