Let $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$, where $a,b,c$ are constants. represent a circle passing through the point $\left(2,5\right)$. Then the shortest distance of the point $\left(11,6\right)$ from this circle is

Let equation of circle is

$x^2+y^2+2gx+2fy+c=0$

Differentiating above equation w.r.t $x$ we get, 

$\Rightarrow \frac{dy}{dx}=-\frac{\left(2x+2g\right)}{\left(2y+2f\right)}$

Now, comparing with $\frac{dy}{dx}=\frac{ax-by+a}{bx+cy+a}$

We get, $b=0,a=-2,c=2$

$\Rightarrow -2g=-2\Rightarrow g=1$

Also, $2f=-2$

So, $f=-1$

Now circle will be

$x^2+y^2+2x-2y+c=0$

its passes through $\left(2,5\right)$

which will give $c=-23$

so circle will be $x^2+y^2+2x-2y-23=0$

centre $C=\left(-1,1\right)$

and radius $5$

Now $P$ is $\left(11,6\right)$

So minimum distance of $P$ from circle will be $=\sqrt{{\left(11+1\right)}^2+{\left(6-1\right)}^2}-5$

$=13-5=8$