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# If the lines $2x+3y+1=0,\qquad 3x-y-4=0$ lie along diameter of a circle of circumference $10\pi$ then the equation of the circle is :

$\begin{array}{1 1}(A)\;x^2+y^2+2x-2y-23=0\\(B)\;x^2+y^2-2x-2y-23=0 \\(C)\;x^2+y^2+2x+2y-23=0 \\(D)\;x^2+y^2-2x+2y-23=0 \end{array}$

Two diameters are along
$2x+3y+1=0$
$3x-y-4=0$
on solving we get centre as $(1,-1)$
We know that circumference $=2 \pi r$
$2 \pi r=10 \pi$
$2r= 10$
$r=5$
Required circle is $(x-1)^2+(y+1)^2=5^2$
$x^2+y^2-2x+2y-23=0$
Hence D is the correct answer.