Consider a circle C: $x^2+y^2-{\alpha }^{2}x-(4\alpha -4)y+c=0$ and a parabola $P:f(x,y)=0,$ if the two curves are orthogonal to each other at $A(2,3)$ and $B(3,2)$ , then sum of radius of circle and latus rectum of parabola is:

Clearly, centre of circle lies on perpendicular bisector of AB i.e, it lies on y=x 
 $\therefore \frac{{\alpha }^2}{2}=\frac{(4\alpha -4)}{2}$
 ${\alpha }^2-4\alpha +4=0$
 $\alpha =2$
$\therefore$ centre (2,2)
 $r=\sqrt{{(2-2)}^2+{(3-2)}^2}=1$
Also, equation of tangents at A and B are x=2, y=2 respectively, which are perpendicular and equal in length.
$\Rightarrow y=x$ is axis of parabola and centre of circle is foot of directrix with AB as latus rectum
 $\Rightarrow AB=\sqrt{2}$
 $S(\frac{5}{2},\frac{5}{2}),vertex(\frac{9}{4},\frac{9}{4})$
Equation of tangent at vertex is  $x+y=\frac{9}{2}$