If the lines $2 x-3 y=5$ and $3 x-4 y=7$ are the diameters of a circle of area 154 sq units, then equation…
If the lines $2 x-3 y=5$ and $3 x-4 y=7$ are the diameters of a circle of area 154 sq units, then equation of the circle is $\left(\right.$ Take $\left.\pi=\frac{22}{7}\right)$
$x^2+y^2-2 x+2 y-47=0$
$x^2+y^2-2 x+2 y-49=0$
$x^2+y^2-2 x-2 y-47=0$
$x^2+y^2-2 x-2 y-49=0$
Solution
Centre is point of intersection of $2 x-3 y=5$ and $3 x-4 y=7$
$i \cdot e \cdot(1,-1)$
and radius $r$ is such that $\pi r^2=154 \Rightarrow r=7$
Hence, the required equation is
$\begin{aligned} & (x-1)^2+(y+1)^2=7^2 \\ \Rightarrow & x^2+y^2-2 x+2 y-47=0\end{aligned}$