Multiple choice

The lengths of tangents from two given points to a given circle are $\displaystyle t_{1}: &: t_{2}$. If the two points are conjugate to each other w.r.to the given circle the distance between the points will be

  1. $\displaystyle \sqrt{\dfrac{t_{1}^{2}+t_{2}^{2}}{2}}$
  2. $\displaystyle \sqrt{t_{1}^{2}+t_{2}^{2}}$
  3. $\displaystyle \sqrt{t_{1}t_{2}}$
  4. $\displaystyle \dfrac{2t_{1}t_{2}}{t_{1}+t_{2}}$
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B Correct answer
Explanation

For conjugate points with respect to a circle, the distance d between them satisfies d^2 = t1^2 + t2^2, where t1 and t2 are the lengths of the tangents from the points to the circle.

AI explanation

Let the circle be x^2 + y^2 = a^2, and let the coordinates of the two points be (x1, y1) and (x2, y2). Since the points are conjugate to each other with respect to the circle, the polar of one passes through the other, giving the relation x1*x2 + y1*y2 = a^2. The lengths of the tangents from these points are t1^2 = x1^2 + y1^2 - a^2 and t2^2 = x2^2 + y2^2 - a^2. The distance d between the two points is given by d^2 = (x1 - x2)^2 + (y1 - y2)^2 = (x1^2 + y1^2 - a^2) + (x2^2 + y2^2 - a^2) - 2(x1*x2 + y1*y2) + 2a^2. Substituting the conjugate relation and tangent expressions, we get d^2 = t1^2 + t2^2 - 2a^2 + 2a^2, which simplifies to d^2 = t1^2 + t2^2. Therefore, the distance between the points is the square root of (t1^2 + t2^2).

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