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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 :

Option 1)
(12, 32)

Option 2)
(18, 42)

Option 3)
(12, 24)

Option 4)
(18, 48)

Answers (1)

best_answer

As we learnt in

Common tangents of two circle -

When they intersect, there are two common tangents, both of them being direct.

- wherein

$S_{1}:x^{2}+y^{2}-4x-4y+6=0$

$r_{1}\:=\:\sqrt{2^{2}+2^{2}-6}\:=\sqrt{2}$

$C_{1}:\:(2,2)$

$S_{2}:\:x^{2}+y^{2}-10x-10y+\lambda=0$

$r_{2}\:=\sqrt{5^{2}+5^{2}-\lambda}\:=\:\sqrt{50-\lambda}$

$C_{2}:\:(5,5)$

For two tangents

$r_{1}-r_{2}< C_{1}C_{2}< \:r_{1}+r_{2}$

$\sqrt{50-\lambda}-\sqrt{2}< 3\sqrt{2}< \sqrt{2}+\sqrt{50-\lambda}$

We get $\sqrt{50-\lambda}< 4\sqrt{2}\; \Rightarrow \lambda> 18$

$\sqrt{50-\lambda}> 2\sqrt{2}\; \Rightarrow \lambda< 42$

$\lambda\epsilon (18,42)$

Option 1)

(12, 32)

This option is incorrect.

Option 2)

(18, 42)

This option is correct.

Option 3)

(12, 24)

This option is incorrect.

Option 4)

(18, 48)

This option is incorrect.

Posted by

divya.saini

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