Question

A variable line passes through the fixed point $(\alpha ,\beta ).$ The locus of the foot of the perpendicular from the origin on the line is,

• A. $x^2+y^2-\alpha x-\beta y=0$
• B. $x^2-y^2+2\alpha x+2\beta y=0$
• C. $\alpha x+\beta y\pm \sqrt{({\alpha }^2+{\beta }^2})=0$
• D. $\frac{x^2}{{\alpha }^2}+\frac{y^2}{{\beta }^2}=1$

Answer 1

The correct option is A $x^2+y^2-\alpha x-\beta y=0$

Let foot of perpendicular from origin on variable line be $P$.
Line joining $(0,0)$ and $(\alpha ,\beta )$ always subtends right angle at $P$.

Thus, Locus of $P$ is a circle with $(0,0)$ and $(\alpha ,\beta )$ as end points of diameter.

Radius $=\frac{\sqrt{{\alpha }^2+{\beta }^2}}{2}$
Centre $\equiv (\frac{\alpha }{2},\frac{\beta }{2})$

Hence, the equation of the circle is ${(x-\frac{\alpha }{2})}^2+{(y-\frac{\beta }{2})}^2=\frac{{\alpha }^2+{\beta }^2}{4}$
$\Rightarrow x^2+y^2-\alpha x-\beta y=0$