Question

The locus of the foot of the perpendicular, from the origin to chords of the circle $x^2+y^2-4x-6y-3=0$ which subtend a right angle at the origin, is

• A. $2(x^2+y^2)-4x-6y-3=0$
• B. $x^2+y^2+4x+6y+3=0$
• C. $2(x^2+y^2)+4x+6y-3=0$
• D. None of these

Answer 1

The correct option is A $2(x^2+y^2)-4x-6y-3=0$

The equation to one such chord, on which the foot of the perpendicular from the origin is $(x_1,y_1)$ is $x{x}_1+y{y}_1={x_1}^2+{y_1}^2$.
Using this homogenise $x^2+y^2-4x-6y-3=0$
This gives $(x^2+y^2){({x_1}^2+{y_1}^2)}^2-2(2x+3y)(x{x}_1+y{y}_1)({x_1}^2+{y_1}^2)-3(x{x}_1+y{y}_1)=0$
These two lines are at right angles
$\therefore 2{({x_1}^2+{y_1}^2)}^2-2(2x_1+3y_1)({x_1}^2+{y_1}^2)-3({x_1}^2+{y_1}^2)=0$
Since ${x_1}^2+{y_1}^2\ne 0$
Hence locus of $(x_1,y_1)$ is $2(x^2+y^2)-2(2x+3y)-3=0$