Find the locus of the foot of the perpendicular from the origin to chord of the circle $x^{2} + y^{2} - 4 x - 6 y - 3 = 0$ , which subtend a right angle at the origin.

2 x^{2} + 2 y^{2} - 4 x + 6 y - 3 = 0

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2 x^{2} + 2 y^{2} + 4 x - 6 y - 3 = 0

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2 x^{2} + 2 y^{2} - 4 x + 6 y + 3 = 0

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2 x^{2} + 2 y^{2} - 4 x - 6 y - 3 = 0

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Solution

The correct option is D $2 x^{2} + 2 y^{2} - 4 x - 6 y - 3 = 0$

Let point be $P (h, k)$

Slope of perpendicular $O P$ is $\frac{k}{h}$

Slope of chord $A B$ is $- \frac{h}{k}$

Equation of line is $(y - k) = - \frac{h}{k} (x - h)$

$k y - k^{2} + h x - h^{2} = 0$

$\frac{k y + h x}{h^{2} + k^{2}} = 1$ ---- (1)

Homogenise the equtaion of circle by (1),

$x^{2} + y^{2} - 4 x (\frac{k y + h x}{h^{2} + k^{2}}) - 6 y$

$\frac{k y + h x}{h^{2} + k^{2}} - 3 \frac{k y + h x}{h^{2} + k^{2}} = 0$

As it subtends a right angle at the origin.

Coefficient of $x^{2}$ + coeffiecient of $y^{2}$ = 0

$(1 - \frac{4 h}{h^{2} + k^{2}} - \frac{3 h^{2}}{(h^{2} + k^{2})^{2}}) + (1 - \frac{6 k}{h^{2} + k^{2}} - \frac{3 k^{2}}{(h^{2} + k^{2})^{2}}) = 0$

$2 - \frac{2 (2 h + 3 k)}{h^{2} + k^{2}} - \frac{3}{h^{2} + k^{2}} = 0$

$2 (h^{2} + k^{2}) - 4 h - 6 k - 3 = 0$

So the locus is $2 x^{2} + 2 y^{2} - 4 x - 6 y - 3 = 0$

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