If one common tangent of two circles x^2 + y^2 = 4 and {x^2} + {left( {y - 3} right)^2} = λ ,

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10-1.Circle and System of Circles

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If one common tangent of the two circles $x^2 + y^2 = 4$ and ${x^2} + {\left( {y - 3} \right)^2} = \lambda ,\lambda > 0$ passes through the point $\left( {\sqrt 3 ,1} \right)$, then possible value of $\lambda$ is

A

$\frac{1}{4}$

B

$\frac{1}{2}$

C

$1$

D

$2$

Solution

Common tangent $\mathrm{T}=0$

$\sqrt{3} x+y=4$

Use $[\perp \text { distance from }(0,3)]^{2}=\lambda$

$\left|\frac{1}{2}\right|^{2}=\lambda \Rightarrow \lambda=\frac{1}{4}$

Standard 11

Mathematics

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