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

The centre of the given circle is $O(4,-3)$

The circumcircle of $\Delta PAB$will circumscribe the quadrilateral $PBOA$also. Hence, one of the diameters must be $OP$

So, the equation of circumcircle of $\Delta PAB$will be

$(x-2)(x-4)+(y-3)(y+3)=0$

Or  $x^2+y^2-6x-1=\mathrm{0.....}\left(i\right)$

Director circle of the given ellipse will be

${(x+5)}^2+{(y-3)}^2=9+b^2$

Or  $x^2+y^2+10x-6y+25-b^2=\mathrm{0....}\left(ii\right)$

So, from  $\left(i\right)\text{ and }\left(ii\right)$ by applying the condition of orthogonally, we get

$2[-3\left(5\right)+0(-3)]=-1+25-b^{2}or-30=24-b^2$

Hence $b^2=54$