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Question


The length of the direct common tangent of the circles $$x^{2}+y^{2}-4x-10y+28=0$$ and $$x^{2}+y^{2}+4x-6y+4=0$$ is


A
2
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B
4
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C
11
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D
16
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Solution

The correct option is A $$\sqrt{11}$$
From point (12)
length of direct common tangent $$=\sqrt{d^{2}-(r_{1}-r_{2})^{2}}$$
distance between centres $$d=\sqrt{(4)^{2}+(2)^{2}}$$
$$=\sqrt{20}$$
and $$r_{1}=1$$ and $$r_{2}=4$$
$$\therefore L=\sqrt{20-(3)^{2}}$$
$$L=\sqrt{11}$$

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