The graph of $x^2+y^2+6x-24y+72=0$ and $x^2-y^2+6x+16y-46=0$ intersect at four points. Sum of the distance of these four points from point $(-3,2)$ be $\alpha ,$ then $\frac{\alpha }{20}$ =_____________.

$x^2+y^2+6x-24y+72=0\dots \dots .\left(i\right)$

$x^2-y^2+6x+16y-46=0\dots \dots \dots \left(ii\right)$

(i)+ (ii) , $2x^2+12x-8y+26=0, x^2+6x-4y+13=0$

${(x+3)}^2=4(y-1) \dots \dots \dots \left(iii\right)$

Pt of intersecting also satisfy equation………(iii)

$\sum _{i=1}^4\sqrt{{(xi+3)}^2+{(yi-2)}^2}=\sum _{i=1}^4\sqrt{4(yi-1)+{(yi-2)}^2}\sum \left|yi\right|$

Similarly equation also satisfy $2y^2-40y+118=0, y^2-20y+59=0\Rightarrow \sum _{i=1}^{4}yi=2\times 20=40$