Minimal Distance

Completing the square, we have $(x-14)^2+(y+20)^2=576=24^2,$ so the center $O$ of the circle is $(14, -20)$ and its radius $R=24$ . Let $P=(x,y)$ be a point on set $S$ . Since $P$ lies outside $\Gamma$ and satisfies $PA\cdot PB=100$ , by Power of a Point, we have $PA\cdot PB = PO^2 - R^2\implies PO^2 = 576+100=676=26^2.$ It follows that the set $S$ (that is, the locus of all points $P$ ) is a circle $\Omega$ centered at $O$ and with radius $26$ . Now, if $M$ lies on $\Gamma$ and $N$ is on $S$ (that is, $N$ lies on $\Omega$ ), the minimum length of segment $MN$ is achieved when $O, M, N$ are collinear in that order. Hence, the minimum length of $MN$ is $MN=|OM-ON|=|24-26|=\boxed{2}$ .

$It~is~clear~that~the~point~ will~ be~ nearest~ when~ AB ~passes ~through~ the~ center~ of ~the ~circle~ and ~furthest~ if~ A=B, ~that~ is~ AP~ is~ tangent.\\ Circle~has~a~radius~r =24. ~Let~X=PA~ is~ nearest,~then\\ PA*PB=X*(X+2*24)=100.\\ Solving~quadratic~in ~X, ~X>0,~we~get~X= \Large~~\color{#D61F06}{2}.$

I,at the very beginning, took the special case that PB passes from the center of the given circle.

Let T be a point on the circle such that PT is tangent to it at T .Let r denote the radius, x denote PA ,and t denote the length of PT . [It could have been nice if I could draw a figure here]

Then I got a right triangle whose hypotenuse is r+x and the other two sides r and t .

Since r can be found from the equation of the circle and t from Power of a point, Using Pythagoras Theorem got the value of x .