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(D) Let the coaxial system of circles be given by $x^2 + y^2 + 2g_i x + c = 0$ for $i = 1, 2, 3$,where $g_i$ are variables and $c$ is a constant.Let t

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(D) Let the coaxial system of circles be given by $x^2 + y^2 + 2g_i x + c = 0$ for $i = 1, 2, 3$,where $g_i$ are variables and $c$ is a constant.Let the fixed point be $(h, k)$. The square of the length of the tangent from $(h, k)$ to the $i$-th circle is $t_i^2 = h^2 + k^2 + 2g_i h + c$.The centers of the circles are $P(-g_1, 0)$,$Q(-g_2, 0)$,and $R(-g_3, 0)$.The distances between the centers are $QR = |g_3 - g_2|$,$RP = |g_1 - g_3|$,and $PQ = |g_2 - g_1|$.Consider the expression $QRt_1^2 + RPt_2^2 + PQt_3^2 = (g_3 - g_2)(h^2 + k^2 + 2g_1 h + c) + (g_1 - g_3)(h^2 + k^2 + 2g_2 h + c) + (g_2 - g_1)(h^2 + k^2 + 2g_3 h + c)$.Expanding this,we get $(h^2 + k^2 + c)(g_3 - g_2 + g_1 - g_3 + g_2 - g_1) + 2h(g_1 g_3 - g_1 g_2 + g_2 g_1 - g_2 g_3 + g_3 g_2 - g_3 g_1) = 0 + 0 = 0$.

The lengths of tangents from a fixed point to three circles of a coaxial system are $t_1, t_2, t_3$. If $P, Q,$ and $R$ are the centers of these circles,then $QRt_1^2 + RPt_2^2 + PQt_3^2$ is equal to

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