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Question

What is the length of common tangents of the circles $$x^{2}+y^{2}=6x, x^{2}+y^{2}+2x=0$$ 


A
3
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B
3,33
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C
23
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D
23,33
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Solution

The correct option is B $$2\sqrt{3}$$
$$x^{2}+y^{2}-6x=0$$
$$C_{1}=(3,0), r_{1}=3$$
and $$x^{2}+y^{2}+2x=0$$
$$C_{2}=(-1,0), r_{2}=1$$
$$\Rightarrow r_{1}+r_{2}=4=C_{1}C_{2}=4$$
So, this two circles touching each other externally.
As we know, point of intersection of common tangent divides segment of line joining centre in ration of there radius.
So, $$h=\frac{3\times (-1)-1(3)}{3-1}=-3$$
and $$k=\frac{3\times (0)-1\times (0)}{3-1}=0$$
So, $$P(-3,0)$$ from P length of external common tangent from point $$(1,2)$$
$$=\sqrt{d^{2}-(r_{1}-r_{2})^{2}}$$
$$=\sqrt{4^{2}-2^{2}}$$
$$=\sqrt{12}$$
$$=2\sqrt{3}$$

58027_31383_ans_aac7548620074504bb23c748f14aa370.png

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