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

Let equation of circle is 

$\begin{array}{l}{\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0\\ \\ \Rightarrow \frac{dy}{dx}=\frac{-(2x+2g)}{(2y+2f)}\end{array}$

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

$$\begin{gathered} \begin{array}{l}\Rightarrow \mathrm{b}=0,\mathrm{a}=-2,\mathrm{c}=2\\ \Rightarrow -2 \mathrm{g}=-2\Rightarrow \mathrm{g}=1 \\ 2\mathrm{f}=-2 \Rightarrow f=-1\\ \end{array} \end{gathered}$$
Now circle will be 

${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-2\mathrm{y}+\mathrm{c}=0$

its passes through (2, 5)

which will give $c=-23$

so circle will be ${\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{x}-2\mathrm{y}-23=0$

centre $\mathrm{C}=(-1, 1)$

and radius 5

Now P is (11, 6)

So minimum distance of $\mathrm{P}$ from circle will be

$\begin{array}{l}=\sqrt{{(11+1)}^2+{(6-1)}^2}-5 \\ =13-5 \\ =8 \end{array}$