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Point Form of Tangent: Ellipse

The lengths o...

Question

The lengths of tangents from a fixed point to three circles of a co-axial system are $t_{1}, t_{2}, t_{3}$ respectively. If $P, Q, R$ be the centres of the three circles, show that $Q R t_{1}^{2} + R P t_{2}^{2} + P Q t_{3}^{2} = 0$ .

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Solution

By special choice of axes the equation of coaxial system of circles can be put in the form
$x^{2} + y^{2} + 2 g_{r} x + c = 0$ where $r = 1, 2, 3$ .
The co-ordinates of the centres of the three circles are
$P (- g_{1}, 0), Q (- g_{2}, 0), R (- g_{3}, 0)$
$∴ Q R = (- g_{3}) - (- g_{2}) = g_{2} - g_{3}$
$R P = g_{3} - g_{1}$ and $P Q = g_{1} - g_{2}$ .
If the fixed point from where the tangents be drawn be $(h, k)$ , then
$t_{1}^{2} = h^{2} + k^{2} + 2 g_{1} h + c$ etc.
$∴ Q R t_{1}^{2} + R P t_{2}^{2} + P Q t_{3}^{2}$
$= (g_{2} - g_{3}) {h^{2} + k^{2} + c + 2 g_{1} h} + . . . . + . . . . = 0$
or $(h^{2} + k^{2} + c) \sum (g_{2} - g_{3}) + 2 h \sum g_{2} (g_{1} - g_{3}) = 0$
$∵ \sum (g_{2} - g_{3}) = 0$ and $\sum g_{1} (g_{2} - g_{3}) = 0$ .

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