The lines $2x-3y-5$ and $3x-4y=7$ are the diameters of a circle of area 154 square units. An equation of this circle is $(\mathrm{\pi }=22/7)$

The center of the circle is the point of intersection of the given diameters $2x-3y-5$ and $3x-4y=7$, which is (1, 1) and if $r$ is the radius of the circle, then

 $$\begin{gathered} \pi r^2=154 \\ \Rightarrow r^2=154\times \frac{7}{22} \\ \Rightarrow r=7 \end{gathered}$$.

Hence an equation of the required circle is $$\begin{gathered} (x-1{)}^2+(y+1{)}^2=7^2 \\ \Rightarrow x^2+y^2-2x+2y=47. \end{gathered}$$