Question

$px+qy=40$ is a chord of minimum length of the circle $(x-10{)}^2+(y-20{)}^2=729.$ If the chord passes through $(5,15),$ then $p^{2019}+q^{2019}$ is equal to

• A. $0$
• B. $2$
• C. $2^{2019}$
• D. $2^{2020}$

Answer 1

The correct option is D $2^{2020}$

Given, equation of circle
$(x-10{)}^2+(y-20{)}^2=729$
Centre $\left(C\right)\equiv (10,20)$ and radius $=\sqrt{729}=27$


Equation of chord : $px+qy=40$
Slope of chord $=-\frac{p}{q}$

Since, chord is of minimum length.
$\therefore P(5,15)$ will be the mid point of given chord.
We know that, any chord and the straight line passing through its midpoint and centre of the circle are perpendicular to each other.
$\Rightarrow m_{CP}\times m_{\text{chord}}=-1\Rightarrow \frac{20-15}{10-5}\times \frac{-p}{q}=-1\Rightarrow p=q$

Also, chord passes through $(5,15).$
$\therefore 5p+15p=40$
$\Rightarrow p=q=2$

$\therefore p^{2019}+q^{2019}=2^{2020}$