If the line $3 x-4 y-7=0$ and $2 x-3 y-5=0$ pass through diameters of a circle of area $49 \pi$ square units…
If the line $3 x-4 y-7=0$ and $2 x-3 y-5=0$ pass through diameters of a circle of area $49 \pi$ square units, then the equation of the circle is
$x^2+y^2-2 x+2 y-47=0$
$x^2+y^2+2 x-2 y-51=0$
$x^2+y^2-2 x+2 y+51=0$
$x^2+y^2+2 x+2 y+47=0$
Solution
Centre of the circle is the point of intersection of the diameters
$3 x-4 y-7=0$ and $2 x-3 y-5=0$
which is $(1,-1)$ and $r=7$
$\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ & \Rightarrow x^2+y^2-2 x+2 y-47=0\end{aligned}$