Multiple choice

If $x=3$ is the chord of contact of the circle $x^{2}+y^{2}=81$, then the equation of the corresponding pair of tangents, is

  1. $x^{2}-8y^{2}+54x+729=0$
  2. $x^{2}-8y^{2}-54x+729=0$
  3. $x^{2}-8y^{2}-54x-729=0$
  4. $x^{2}-8y^{2}=729$
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B Correct answer
Explanation

For a circle x^2 + y^2 = r^2, the chord of contact for a point (x1, y1) is xx1 + yy1 = r^2. Here, the chord is x = 3, so x(3) = 81 implies x = 27. The point is (27, 0). The pair of tangents is given by SS1 = T^2, where S = x^2 + y^2 - 81 and T = 27x - 81. Substituting and simplifying yields x^2 - 8y^2 - 54x + 729 = 0.

AI explanation

Using the equation of the pair of tangents drawn from an external point (x1, y1) to a circle, we apply the formula T^2 = S * S1. For the given circle x^2 + y^2 = 81 and the line x = 3, the point of intersection of the tangents on the x-axis is found by xx1 = r^2, giving x1 = 27, so the point is (27, 0) and S1 becomes 648. Substituting into the formula gives (27x)^2 = (x^2 + y^2 - 81)(648), which simplifies to x^2 - 8y^2 - 54x + 729 = 0.

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