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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The center of circle inscribed in square formed by the lines $x^2-8x+12=0$ and $y^2-14\;y+45=0$ is

$\begin{array}{1 1}(A)\;(4,7) \\(B)\;(7,4)\\(C)\;(9,4) \\(D)\;(4,9) \end{array}$

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1 Answer

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$x^2-8x+12=0$
=> $(x-6)(x-2)=0$
$y^2-14y+45=0$
=> $(y-5)(y-9)=0$
Thus sides of square are
$x=2,x=6,y=5,y=9$
Then centre of circle inscribed in square will be $ \bigg( \large\frac{2+6}{2}, \frac{5+9}{2}\bigg)$
=> $ \bigg( \large\frac{8}{2},\frac{14}{2}\bigg)$
=> $(4,7)$
Hence A is the correct answer.
answered Apr 8, 2014 by meena.p
 

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