If $C_{i}$ is the centre of the circle $x^{2}+y^{2}+2g_{i}x+5=0$ and $t_{i}$ is the length of the tangent from any point to this circle, $i=1,2,3$ then the points $(g_{1}, t_{1}^{2})$ , $(g_{2}, t_{2}^{2})$ and $(g_{3}, t_{3}^{2})$ are

Let the point from which the tangent is drawn be (x1, y1). The length squared of the tangent, t_i squared, is found by substituting the point into the circle's equation, giving t_i squared equals x1 squared plus y1 squared plus 2 times g_i times x1 plus 5. Rearranging this equation gives t_i squared equals 2 times x1 times g_i plus (x1 squared plus y1 squared plus 5). This equation is of the form Y equals mX plus c, where X is g_i and Y is t_i squared, proving that all points (g_i, t_i squared) lie on the straight line Y equals 2x1 X plus (x1 squared plus y1 squared plus 5). Therefore, the points are collinear.

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