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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The centre of the circle passes through $(0,0)$ and $(1,0)$ and touching the circle $x^2+y^2=9$ is

$\begin{array}{1 1}(A)\;\bigg(\frac{1}{2},\frac{1}{2}\bigg) \\(B)\;\bigg(\frac{1}{2},-\sqrt 2\bigg) \\(C)\;\bigg(\frac{3}{2},\frac{1}{2}\bigg) \\(D)\;\bigg(\frac{1}{2},\frac{3}{2}\bigg) \end{array}$

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1 Answer

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Let the requirement circle be
$x^2+y^2+2gx+2fy+c=0$
Since it passes through $(0,0)$ and $(1,0)$
=> $C=0$ and $g=\large\frac{-1}{2}$
Points $(0,0)$ and $(1,0)$ lie inside the circle $x^2+y^2=9$ So two circles touch internally
=> $C_1C_2= r_1-r_2$
$\sqrt {g^2+f^2}=3 -\sqrt {g^2+f^2}$
=> $ \sqrt {g^2+f^2}=\large\frac{3}{2}$
=> $ f^2=\large\frac{9}{4}-\frac{1}{4}$$=2$
Hence the centres of required circle are $\bigg(\large\frac{1}{2},$$-\sqrt 2\bigg)$
Hence B is the correct answer.
answered Apr 8, 2014 by meena.p
 

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