Prove that the circles $x^{2} + y^{2} + 24 u x + 2 v y = 0$ and $x^{2} + y^{2} + 2 u_{1} x + 2 v_{1} y = 0$ touch each other if $12 u v_{1} = u_{1} v$ .

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Solution

Let $C_{1} (12 u, - v)$ be the centre of circle $x^{2} + y^{2} + 24 u x + 2 v y = 0$ and $C_{2} (u_{1}, v_{1})$ be the centre of the circle $x^{2} + y^{2} + 2 u_{1} x + 2 v_{1} y = 0$ . According to question distance between $C 1$ and $C 2$ is equal do the sum of radius or both the circles.
$\sqrt (- 12 u + u 1)^{2} + (- v + v 1)^{2} = \sqrt (144 u^{2} + v^{2}) + \sqrt u_{1}^{2} + v_{1}^{2}$
squaring both sides, we get
$(- 12 u + u 1)^{2} + (- v + v 1)^{2} = (144 u^{2} + v^{2}) + (u_{1}^{2} + v_{1}^{2}) + 2 \sqrt (144 u^{2} + v^{2}) \sqrt (u_{1}^{2} + v_{1}^{2}) 144 u^{2} + u_{1}^{2} - 24 u u_{1} + v^{2} + v_{1}^{2} - 2 v v_{1} = 144 u^{2} + v^{2} + u_{1}^{2} + v_{1}^{2} + 2 \sqrt (144 u^{2} + v^{2}) \sqrt (u_{1}^{2} + v_{1}^{2}) - 12 u u_{1} - v v_{1} = \sqrt 144 u^{2} + v^{2} \sqrt u_{1}^{2} + v_{1}^{2}$ $
squaring both sides, we get
$(- 12 u u_{1} - v v_{1}^{2}) = (144 u^{2} + v^{2}) \sqrt (u_{1}^{2} + v_{1}^{2})$
$144 u^{2} u_{1}^{2} + v^{2} v_{1}^{2} + 24 u u_{1} v v_{1} = 144 u^{2} u_{1}^{2} + 144 u^{2} v_{1}^{2} + v^{2} u_{1}^{2} + v^{2} v_{1}^{2}$
$24 u u_{1} v v_{1} = 144 u^{2} v_{1}^{2} + v^{2} u_{1}^{2}$
$(12 u v_{1})^{2} - 2^{} (12 u v_{1})^{} (v u_{1}) + (v u_{1})^{2} = 0$
$(12 u v_{1} - v u_{1})^{2} = 0$
$12 u v_{1} - v u_{1} = 0$
$12 u v_{1} = v u_{1}$

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