Tangents drawn from the point P(2, 3) to the circle  $x^2+y^2-8x+6y+1=0$  touch the circle at the points A and B. The circumcircle of the  $\Delta PAB$ cuts the director circle of ellipse $\frac{{(x+5)}^2}{9}+\frac{{(y-3)}^2}{b^2}=1$  orthogonally. Then the value of  $\left(\frac{b^2}{9}\right)$  is ____

Centre of the given circle is O(4, -3)
The circumcircle of  $\Delta PAB$ will circumscribe the quadrilateral PBOS also, hence one of the diameters must be OP.
Equation of circumcircle of  $\Delta PAB$  will be  $(x-2)(x-4)+(y-3)(y+3)=0$
$\Rightarrow x^2+y^2-6x-1=0$        ……………(1)
Director circle of given ellipse will be
${(x+5)}^2+{(y-3)}^2=9+b^2$
$\Rightarrow x^2+y^2+10x-6y+25-b^2=0$     …………………(2)
From (1) and (2) by applying condition of orthogonally, we get  
$2[-3\left(5\right)+0(-3)]=-1+25-b^2$
$\Rightarrow -30=24-b^2$

$b^2=54\Rightarrow \left(\frac{b^2}{9}\right)$  is 6