Given equation of circle,

$x^2-14x+49+y^2-10y+25-74-151=0$

On rearranging we get,

${\left(x-7\right)}^2+{\left(y-5\right)}^2=225$

$\Rightarrow {\left(x-7\right)}^2+{\left(y-5\right)}^2={15}^2$

So, Centre $=\left(7,5\right)$ and Radius $=15$ units

Now putting $x=2$ & $y=-7$ In the clrcle's equation,

${\left(-5\right)}^2+{\left(-12\right)}^2-225<0$

So, $\left(2,-7\right)$ lies inside the circle.

Now distance from $\left(2,-7\right)$ to $\left(7,5\right)$$=\sqrt{{\left({\mathrm{x}}_2-{\mathrm{x}}_1\right)}^2+{\left({\mathrm{y}}_2-{\mathrm{y}}_1\right)}^2}$

$D=\sqrt{\left\{{\left(7-2\right)}^2+{\left(5-\left(-7\right)\right)}^2\right\}}$

$D=\sqrt{\left\{5^2+{12}^2\right\}}=13$

Now shortest distance will be $r-13=15-13=2$

And largest distance will be $r+13=15+13=28$

So the ratio of largest to shortest will be $14:1$