It's algebra or calculus or something else?

Actually,

$\boxed{\sqrt{p^{2} + q^{2} - 2p + 6q + 9}}$

is the length if tangent drawn from the point $(p,q)$ to the circle

$x^{2} + y^{2} - 2x + 6y +9 = 0$

where the point $(p,q)$ lies on the circle

$x^{2} + y^{2} - 8x - 2y + 13 = 0$ .

Now, the maximum and minimum length of tangents drawn from the point $(p,q)$ to the given circle will be $4\sqrt{3}$ and $2\sqrt{2}$ respectively which can be very easily seen by the diagram. Hence, $\boxed{\frac{M}{m} = \sqrt{6}}$ .

Which gives

$4(\frac{M}{m})^{2} = \boxed{24}$ .

Archit's solution is nice . but i am posting my approach

From the second condition we have

(p-4)^2+(q-1)^2 = 4

We use the parametric coordinates as p = 4+2cosx and q = 1+2sinx

Simply plug in the equation asked and we will get the square of the expression as

28+12cosx+16sinx

Now use the max and min value of acosx+bsinx

Simply max value = 48 and min = 8

So answer = 4*48/8 = 24 :)