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 $A B$ 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 \frac{x}{5}+\frac{3 y}{\lambda}=1 $
$\Rightarrow \frac{3 y}{\lambda}=1-\frac{x}{5} $
$\Rightarrow \frac{1}{\lambda}=\frac{1}{3 y}\left(1-\frac{x}{5}\right)\dots$(i)
and equation of line $B P$ is
$\frac{x}{\lambda / 5}+\frac{y}{-3}=1 $
$\Rightarrow \frac{5 x}{\lambda}-\frac{y}{3}=1$
$\Rightarrow \frac{1}{\lambda}=\frac{1}{5 x}\left(\frac{y}{3}+1\right) \dots$(ii)
From Eqs. (i) and (ii),
$\frac{1}{3 y}\left(1-\frac{x}{5}\right) =\frac{1}{5 x}\left(\frac{y}{3}+1\right)$
$\Rightarrow 5 x\left(1-\frac{x}{5}\right) =3 y\left(\frac{y}{3}+1\right)$
$\Rightarrow 5 x-x^{2} =y^{2}+3 y$
$\Rightarrow x^{2}+y^{2}-5 x+3 y =0$
which is a circle.