Multiple choice

The length of tangents from a fixed point to three circles of coaxial system are t1, t2, t3 and if P, Q and R be the centers, then QRt12 + RPt22 + PQt32 is equal to

  1. 1

  2. 2

  3. 3

  4. 0
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D Correct answer
Explanation

For a coaxial system of circles, the power of a point (t^2) relative to the centers follows a linear relationship. The sum QR*t1^2 + RP*t2^2 + PQ*t3^2 equals 0 due to the properties of the radical axis and power of a point.

AI explanation

The fixed point is one of the limiting points of the coaxial system, from which the tangent length to any circle in the system is zero. Thus, t1 = t2 = t3 = 0. Substituting these values into the expression gives QR(0) + RP(0) + PQ(0) = 0. The result is 0.

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