The locus of the foot of the perpendicular from the origin to chords of the circle $\mathrm{x}^{2}+\mathrm{y}^{2}-4\mathrm{x}-6\mathrm{y}-3=0$ which substend a right angle at the origin, is
- $2(\mathrm{x}^{2}+\mathrm{y}^{2})-4\mathrm{x}-6\mathrm{y}-3=0$
- $\mathrm{x}^{2}+\mathrm{y}^{2}-4\mathrm{x}-6\mathrm{y}-3=0$
- $2(\mathrm{x}^{2}+\mathrm{y}^{2})+4\mathrm{x}+6\mathrm{y}-3=0$
- $2(\mathrm{x}^{2}+\mathrm{y}^{2})+4\mathrm{x}+6\mathrm{y}+3=0$
The chord subtends 90 degrees at the origin. If the chord equation is lx + my = 1, the condition for 90 degrees is a^2 + b^2 + 2g(l) + 2f(m) + c = 0. The locus of the foot of the perpendicular (h, k) from origin to the chord is h^2 + k^2 = hx + ky. Substituting into the circle equation leads to the result.
Let the foot of the perpendicular from the origin to the chord be (h, k). The equation of this chord is hx + ky = h^2 + k^2. Since the chord subtends a right angle at the origin, the origin must lie on the circle drawn on the chord as diameter, leading to the combined equation representing pair of lines as h(x^2 + y^2) + k(x^2 + y^2) = (hx + ky)^2 evaluated with the circle's properties. A simpler method uses the fact that for such a chord of the circle S = 0, the locus of the midpoint (which coincides with the perpendicular foot from the origin for right-angled chords) is given by T = S1. Substituting the origin (0,0) as the point (x1, y1) into the circle x^2 + y^2 - 4x - 6y - 3 = 0, we apply the transformation T = S1 to get hx - 2x - 3y = h^2 + k^2, which simplifies to the locus equation 2(x^2 + y^2) - 4x - 6y - 3 = 0.