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 substend 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. $2(x^2+y^2)+4x+6y+3=0$

Answer 1

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

Let the foot of perpendicular from origin to a chord of circle $x^2+y^2-4x-6y-3=0$ has coordinates $(h,k)$. Hence, slope of the line joining origin and foot of perpendicular is $\frac{k}{h}$. So, slope of the chord is $-\frac{h}{k}$.
Equation of the chord is $y-k=-\frac{h}{k}(x-h)$
$hx+ky=h^2+k^2$.....(1)
Now, the combined equation of the two lines joining the origin and two ends of the chord is obtained by homogenizing the equation of circle with the help of (1).
$x^2+y^2+(-4x-6y)\left(\frac{hx+ky}{h^2+k^2}\right)-3{\left(\frac{hx+ky}{h^2+k^2}\right)}^2=0$
Since the two lines are perpendicular (given that they subtend right angle at origin), in the combined equation, coefficient of $x^2$ + coefficient of $y^2=0$
$\Rightarrow [1-\frac{4h}{h^2+k^2}-\frac{3h^2}{{(h^2+k^2)}^2}]+[1-\frac{6k}{h^2+k^2}-\frac{3k^2}{{(h^2+k^2)}^2}]=0$
$\Rightarrow 2(h^2+k^2)-(4h+6k)-3=0$, which is the desired locus.
$\Rightarrow 2(x^2+y^2)-(4x+6y)-3=0$, which is the required locus.
Hence, option 'A' is correct.