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The set of all real values of $$\lambda$$ for which exactly two common tangents can be drawn to the circles
$$x^2+y^2-4x-4y+6=0$$ and $$x^2+y^2-10x-10y+\lambda=0$$ is the interval :


A
(18,42)
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B
(12,32)
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C
(12,24)
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D
(18,48)
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Solution

The correct option is A $$(18, 42)$$
Two circles can have exactly two common tangents only if the circles intersect each other. For the circles to intersect each other, the following conditions must hold true - 
$$r_1+r_2  >  dist (c_1, c_2)$$
$$r_2 - r_1  < dist (c_1, c_2)$$
$$\sqrt{2} + \sqrt{50 -\lambda}  > 3 \sqrt{2} \Rightarrow 50 - \lambda< 8 \Rightarrow \lambda < 42$$
$$\sqrt{50 - \lambda} - \sqrt{2} < 3\sqrt{2} \Rightarrow 50- \lambda  < 32 \Rightarrow \lambda > 18$$
$$\therefore\ \lambda$$ lies in the interval $$(18,24)$$ 

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