Question

A line cuts the $x$-axis at $A(5,0)$ and the $y$-axis at $B(0,-3)$. A variable line $PQ$ is drawn perpendicular to $AB$ cutting the $x$-axis at $P$ and the $y$-axis at $Q$. If $AQ$ and $BP$ meet at $R$, then the locus of $R$ is

• A. $x^2+y^2-5x+3y=0$
• B. $x^2+y^2+5x+3y=0$
• C. $x^2+y^2+5x-3y=0$
• D. $x^2+y^2-5x-3y=0$

Answer 1

Answer: (A)

Equation of line $AB$ is
$\frac{x}{5}+\frac{y}{-3}=1$
$\Rightarrow 3x-5y=15$

Perpendicular line to $AB$ is
$5x+3y=\lambda$
Coordinate of $P$ is $\left(\frac{\lambda }{5},0\right)$
and coordinate of $Q$ is $(0,\lambda /3)$
Now, equation of line $AQ$ is
$x/5+\frac{y}{\lambda /3}=1$
$\Rightarrow \frac{x}{5}+\frac{3y}{\lambda }=1$
$\Rightarrow \frac{3y}{\lambda }=1-\frac{x}{5}$
$\Rightarrow \frac{1}{\lambda }=\frac{1}{3y}\left(1-\frac{x}{5}\right)\dots$(i)
and equation of line $BP$ is
$\frac{x}{\lambda /5}+\frac{y}{-3}=1$
$\Rightarrow \frac{5x}{\lambda }-\frac{y}{3}=1$
$\Rightarrow \frac{1}{\lambda }=\frac{1}{5x}\left(\frac{y}{3}+1\right)\dots$(ii)
From Eqs. (i) and (ii),
$\frac{1}{3y}\left(1-\frac{x}{5}\right)=\frac{1}{5x}\left(\frac{y}{3}+1\right)$
$\Rightarrow 5x\left(1-\frac{x}{5}\right)=3y\left(\frac{y}{3}+1\right)$
$\Rightarrow 5x-x^2=y^2+3y$
$\Rightarrow x^2+y^2-5x+3y=0$
which is a circle.