Question

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

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

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 $$(1,2)$$$$=\sqrt{d^{2}-(r_{1}-r_{2})^{2}}$$$$=\sqrt{4^{2}-2^{2}}$$$$=\sqrt{12}$$$$=2\sqrt{3}$$Maths

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