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Find the locus of a point from which the lengths of the tangents to the circles $x^2+y^2+4x+3=0$ and $x^2+y^2-6x+5=0$ are in the ratio $2:3$.

My Attempt:

Given circles are: $$x^2+y^2+4x+3=0$$ and, $$x^2+y^2-6x+5=0$$. Let, $P(x_1,y_1)$ be a point on the locus, and $T_1$ and $T_2$ be the lengths of tangents.

According to question: $T_1:T_2=2:3$ $\frac {T_1}{T_2}=\frac {2}{3}$ $3T_1=2T_2$.

Please help me to continue from here.

pi-π
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    You have not tried anything. You have just restated what is given.. Do you know the equation of a tangent from an outside point? Show us what you have done.! – Qwerty Oct 09 '16 at 18:38
  • @Qwerty, I have shown, what I have done. I do not know, how to find the length of the tangents, or the equation of a tangent from an external point. – pi-π Oct 10 '16 at 01:49
  • Study coordinate geometry first, especially the chapter on Circles. Then try to solve.. – Qwerty Oct 10 '16 at 06:56
  • Hint:- $T_1 = \sqrt (x_1^2 + y_1^2 + 4x_1 + 3)$ – Mick Oct 11 '16 at 03:42

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