Let dy dx= ax-by+a bx+cy+a, where a,b,c are constants. represent a circle passing through the point (2,5). Then the shortest distance of the point…

Let equation of circle is

$${x}^{2}+{y}^{2}+2gx+2fy+c=0$$x2+y2+2gx+2fy+c=0

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

$$\Rightarrow \frac{dy}{dx}=-\frac{(2x+2g)}{(2y+2f)}$$⇒dxdy​=−(2y+2f)(2x+2g)​

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

We get, $b=0,a=-2,c=2$b=0,a=−2,c=2

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

Also, $2f=-2$2f=−2

So, $f=-1$f=−1

Now circle will be

$${x}^{2}+{y}^{2}+2x-2y+c=0$$x2+y2+2x−2y+c=0

its passes through $(2,5)$(2,5)

which will give $c=-23$c=−23

so circle will be ${x}^{2}+{y}^{2}+2x-2y-23=0$x2+y2+2x−2y−23=0

centre $C=(-1,1)$C=(−1,1)

and radius $5$5

Now $P$P is $(11,6)$(11,6)

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

$$=13-5=8$$=13−5=8