If $C_{i}$ is the centre of the circle $x^{2} + y^{2} + 2 g_{i} x + 5 = 0$ 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

Collinear

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non collinear

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either collinear or non collinear

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not defined

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Solution

The correct option is A Collinear
Let, $p (h, k)$ is any point
So, $x^{2} + y^{2} + 2 g i x + 5 = 0$
$t_{1}^{2} = h^{2} + k^{2} + 2 g_{1} h + 5$
$A (g_{1}, h^{2} + k^{2} + 25_{1} h + 5)$
So, $x^{2} + y^{2} + 2 g_{2} x + 5 = 0$
$t_{2}^{2} = h^{2} + k^{2} + 2 g_{2} h + 5$
$(g_{2}, h^{2} + k^{2} + 2 g_{2} h + 5)$
and $x^{2} + y^{2} + 2 g_{3} x + 5 = 0$
$t_{2}^{2} = h^{2} + k^{2} + 2 g_{3} h + 5$
$(g_{2}, h^{2} + k^{2} + 2 g_{3} h + 5)$
So, stepe of AB = $\frac{2 (g_{1} - g_{2}) h}{g_{1} - g_{2}} = 2 h$
stepe of BC= $\frac{2 (g_{2} - g_{3}) h}{g_{2} - g_{3}} = 2 h$
So, $A, B$ and $C$ are collinear.

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