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  • From a point (h, 0) common tangents are drawn to the circles and

From a point (h, 0) common tangents are drawn to the circles \mathrm{x^2+y^2=1} and
\mathrm{(x-2)^2+y^2=4}, the value of h is

Option: 1

2


Option: 2

– 2


Option: 3

– 2/3


Option: 4

 none of these


Answers (1)

best_answer

Equation of tangent to the circle \mathrm{x^2+y^2=1} is 

\mathrm{y=m x \pm \sqrt{1+m^2}}

This also touches the circle \mathrm{(x-2)^2+y^2=4}

\mathrm{\text { So }\left|\frac{2 m \pm \sqrt{1+m^2}}{\sqrt{1+m^2}}\right|=2 \Rightarrow m= \pm \frac{1}{\sqrt{3}}}

  • common tangent are \mathrm{y=\frac{1}{\sqrt{3}} x+\frac{2}{\sqrt{3}} \text { and } y=-\frac{1}{\sqrt{3}} x-\frac{2}{\sqrt{3}}}, on putting y = 0. From both equations we get x = – 2  ⇒ h = – 2.

 

 

Posted by

Rakesh

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