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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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If a circle passes through the points $(a,b)$ and cuts the circle $x^2+y^2=4$ orthogonally, then the locus of its centre is

$\begin{array}{1 1}(A)\;2ax-2by-(a^2+b^2+4)=0 \\(B)\;2ax-2by-(a^2+b^2+4)=0 \\(C)\;2ax-2by+(a^2+b^2+4)=0 \\(D)\;2ax+2by+(a^2+b^2+4)=0 \end{array}$

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1 Answer

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Let the variable of the circle is :
$x^2+y^2+2gx+2fg+c=0$
It passes through (x,y)
$\therefore \;a^2+b^2+2gx+2fy+c=0$----(i)
It passes through (a,b)
$\therefore \;a^2+b^2+2ga+2fb+c=0$-----(ii)
(i) cuts $x^2+y^2=4$ orthoganally
$\therefore 2(g \times 0 +f \times 0)=c-4$
$=>c=4$
from (ii) $a^2+b^2+2ga+2fb+4=0$
Locus of center $(-g,-f)$ is
$a^2+b^2-2ax+2by+4=0$
or $2ax+2by=a^2+b^2+4$
Hence B is the correct answer.
answered Apr 9, 2014 by meena.p
 

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