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The number of common tangent to the circle $x^2+y^2=4$ and $x^2+y^2-6x-8y=24$

$\begin{array}{1 1}(A)\;0 \\(B)\;1 \\(C)\;3\\(D)\;4 \end{array}$

1 Answer

Given $x^2+y^2=4$
center $C_1=(0,0)$
Also the circle $x^2+y^2-6x-8y-24=0$
$C_2=(3,4)$ and $R_2=7$
Again $C_1C_2=5=> R_2-R_1$
Therefore the given circle touch internally such that they can have just one common tangent at the point of constant .
Hence B is the correct answer.
answered Apr 8, 2014 by meena.p

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