Tangents drawn from P (1, 8) to the circle $x^2+y^2-6x-4y-11=0$ touches the circle at  the points A and B, respectively. The radius of the circle which passes through the points  of intersection of circles  $x^2+y^2-2x-6y+6=0$ and  $x^2+y^2+2x-6y+6=0$ and  intersects the circumcircle of the  $\Delta$ PAB orthogonally is equal to R, then the value of  $4R^2$ is ___

Equations of circle circumscribing $\Delta PAB$  is 
 
$(x-1)(x-3)+(y-8)(y-2)=0$ 
 $\Rightarrow x^2+y^2-4x-10y+19=0$
Equation of circle passing through points of intersection of circles $x^2+y^2-2x-6y+6=0$   and   $x^2+y^2+2x-6y+6=0$  is given by   
 $(x^2+y^2-2x-6y+6)+\lambda (x^2+y^2+2x-6y+6)=0$

$\Rightarrow x^2+y^2+\frac{(2\lambda -2)}{\lambda +1}x-6y+6=0$
As circle (ii) is orthogonal to circle(i), we have
 $$\begin{gathered} -2\left(\frac{2\lambda -2}{\lambda +1}\right)-5(-6)=19+6 \\ \Rightarrow 4\lambda -4=5\lambda +5 \\ \Rightarrow \lambda =-9 \end{gathered}$$
 Hence, required equation of circle is 
 $x^2+y^2+\frac{5}{2}x-6y+6=0$
$\therefore$  Radius of circle  $=\sqrt{\frac{25}{16}+9-6}=\frac{\sqrt{73}}{4}$