A line cuts the $X$ -axis at $A(5,0)$ and the $Y$ -axis at $B(0,-3)$. A variable line $P Q$ is drawn perpendicular to $A B$ cutting the $X$ -axis at $P$ and the $Y$ -axis at $Q$. If $A Q$ and $B P$ meet at $R$, then the locus of $R$ is

• Correct answer
  A

  $x^{2}+y^{2}-5 x+3 y=0$
• B

  $x^{2}+y^{2}+5 x+3 y=0$
• C

  $x^{2}+y^{2}+5 x-3 y=0$
• D

  $x^{2}+y^{2}-5 x-3 y=0$

Solution

Equation of line $A B$ is $$ \begin{array}{l} \frac{x}{5}+\frac{y}{-3}=1 \\ 3 x-5 y=15 \end{array} $$ Perpendicular line to $A B$ is $$ 5 x+3 y=\lambda $$ Coordinate of $P$ is $\left(\frac{\lambda}{5}, 0\right)$ and coordinate of $Q$ is $(0, \lambda / 3)$ Now, equation of line $A Q$ is $x / 5+\frac{y}{\lambda / 3}=1$ $\Rightarrow \quad \frac{x}{5}+\frac{3 y}{\lambda}=1$ $\Rightarrow \quad \frac{3 y}{\lambda}=1-\frac{x}{5}$ $\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{3 y}\left(1-\frac{x}{5}\right)$ and equation of line $B P$ is $\frac{x}{\lambda / 5}+\frac{y}{-3}=1$ $\Rightarrow \quad \frac{5 x}{\lambda}-\frac{y}{3}=1$ $\Rightarrow \quad \frac{1}{\lambda}=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$ From Eqs. (i) and (ii), $\frac{1}{3 y}\left(1-\frac{x}{5}\right)=\frac{1}{5 x}\left(\frac{y}{3}+1\right)$ $\Rightarrow \quad 5 x\left(1-\frac{x}{5}\right)=3 y\left(\frac{y}{3}+1\right)$ $\Rightarrow \quad 5 x-x^{2}=y^{2}+3 y$ $\Rightarrow x^{2}+y^{2}-5 x+3 y=0$ which is a circle.

Hence, the correct answer is option: 'A'.