Multiple choice

The locus of the centre of a circle which passes through the point $(h,k)$ and cuts off a chord of length $2d$ on the line $lx + my + n = 0$ is

  1. $(l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n})^{2}=(l^{2}+\mathrm{m}^{2})[(x-h)^{2}+(y-k)^{2}-d^{2}]$
  2. $(l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n})^{2}=(l^{2}+\mathrm{m}^{2}) [(y-h)^{2}+(x-k)^{2}-d^{2}]$
  3. $(l\mathrm{x}+$ my $+\mathrm{n})^{2}=\mathrm(l^{3}+\mathrm{m}^{3})\lfloor(2x-2h)^{2}+(2y-10k)^{2}-d^{2}\rfloor$
  4. $(l\mathrm{x}+\mathrm{m}\mathrm{y}+\mathrm{n})^{2}=(l^{3}+\mathrm{m}^{3})[(2x-2h)^{2}+(2y-10k)^{2}-d^{2}]$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a circle with centre (x,y) passing through (h,k), the radius squared is (x-h)^2 + (y-k)^2. A chord of length 2d has distance squared r^2 - d^2 from the centre to its line, which gives the equation in option A.

AI explanation

Let the centre of the circle be (x, y) with radius R. Because the circle passes through (h, k) and cuts a chord of length 2d on the line lx + my + n = 0, we have R^2 = (x - h)^2 + (y - k)^2. The perpendicular distance from the centre to the chord line is d = |lx + my + n| / sqrt(l^2 + m^2). Using the geometric relation R^2 = (perpendicular distance)^2 + (half chord length)^2, we substitute the values to get (x - h)^2 + (y - k)^2 = [(lx + my + n)^2 / (l^2 + m^2)] + d^2. Rearranging this equation yields (lx + my + n)^2 = (l^2 + m^2)[(x - h)^2 + (y - k)^2 - d^2].