Question

$px+qy=40$ 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

• A. $0$
• B. $2$
• C. $2^{2013}$
• D. $2^{2014}$

Answer 1

The correct option is D $2^{2014}$


Clearly, $(5,15)$ is the mid-point of the chord.
Hence, slope of normal to the chord is $\frac{20-15}{10-5}=1$
$\therefore$ Slope of chord $=-1$
$\therefore$ Equation of chord is $(y-15)=-(x-5)$
$\Rightarrow x+y=20$
$\Rightarrow 2x+2y=40$
$\therefore p^{2013}+q^{2013}=(2{)}^{2013}+(2{)}^{2013}=2^{2014}$