$px +qy=40$  is a chord of minimum length of circle  ${(x–10)}^2+{(y–20)}^2=729$. If the chord passes through $(5,15)$, then $p^{2013}+q^{2013}$  is equal to

Clearly (5, 15) is the mid point of the chord hence slope of normal to the chords is

$\frac{20-15}{10-5}=1,$ slope of chord = -1. 
Equation of chord (y – 15) =- (x – 5) 
$\Rightarrow 2x+2y=40$$\Rightarrow x+y=20$
$\Rightarrow 2^{2013}+q^{2013}={\left(2\right)}^{2013}+{\left(2\right)}^{2013}=2^{2014}$