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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The lines $2x-3y=5$and $3x-4y=7$ are diameters of a circle having area as $154$ sq.units. The equation of the circle is .

$\begin{array}{1 1}(A)\;x^2+y^2-2x+2y=62 \\(B)\;x^2+y^2+2x-2y=62 \\(C)\;x^2+y^2+2x-2y=47 \\(D)\;x^2+y^2-2x+2y=47 \end{array}$

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1 Answer

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$\pi r^2=154$
$r=7$
For centre on slowing equation we get,
$2x-3y=5$
$3x-4y=7$
$6x-9y=15$
$6x+8y=14$
______________
$\quad \qquad-y=1$
$\quad \qquad y=1$
$\quad \qquad x=1$
So center is (1,-1)
Equation of the circle $(x-1)^2+(y+1)^2=7^2$
=> $x^2+y^2-2x+2y=47$
Hence D is the correct answer.
answered Apr 8, 2014 by meena.p
 

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