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# The equation of the circle circumscribing the triangle whose sides are the lines $y=x+2,3y=4x,2y=3x$ is _____________

$\begin{array}{1 1}x^2+y^2+122x-196y=0\\x^2+y^2-122x-196y=0\\x^2+y^2-122x+196y=0\\x^2+y^2+122x+196y=0\end{array}$

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• General equation of a circle with centre $(g,f)$ is $x^2+y^2+2gx+2fy+c=0$
Answer : $x^2+y^2+122x-196y=0$
Let the equations of the sides $AB,BC$ and $CA$ of the triangle be $y=x+2$----(1),$3y=4x$-----(2) and $2y=3x$----(3) respectively.
On solving (1) & (2) we get,
$3(x+2)=4x$
$\Rightarrow 3x+6=4x$
$\therefore x=6$ and $y=8$
Solving (2) & (3) we get,
$x=0,y=0$
Solving (3) & (1) we get,
$2(x+2)=3x$
$\Rightarrow 2x+4=3x$
$\therefore x=4$ and $y=6$
Hence the vertices of the triangle are $A(4,6),B(6,8)$ and $C(0,0)$
The circle $x^2+y^2+2gx+2fy+c=0$ passes through $(4,6),(6,8)$ and $(0,0)$
$\therefore 8g+12f+c=-52\Rightarrow 2g+3f=-13$
$12g+16f+c=-100\Rightarrow 3g+4f+c=-50$
$c=0$
$2g+3f=-13$----(4)
$3g+4f=-50$----(5)
$\;\;(Eq(4)\times 3)6g+9f=-\;39$
$-(Eq(5)\times 2)6g+8f=-100$
______________________________
$\Rightarrow f=61$
$\therefore g=-98$
$c=0$
Hence the equation of the required circle is $x^2+y^2+122x-196y=0$