Let (dy)/(dx) = (ax - by + a)/(bx + cy + a) , where a, b, c are...

Let equation of circle is

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

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

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

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

We get, $b = 0, a = - 2, c = 2$

$⇒ - 2g = - 2 ⇒ g = 1$

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)$</p><p>which will give$c = - 23$</p><p>so circle will be${x}^{2} + {y}^{2} + 2x - 2y - 23 = 0$</p><p>centre$C = \left(- 1, 1\right)

and radius $5$

Now $P$is \left(11, 6\right)

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

$= 13 - 5 = 8$