If the lines x + 2 y - 5 = 0 and 2 x - 3 y + 4 = 0 lie along diameters of a circle of area 9 π , then…

Given: the lines $x+2y-5=0$ and $2x-3y+4=0$ lie along diameters of a circle of area $9\pi$

We have $\pi r^2=9\mathrm{\pi }\Rightarrow r^2=9$

The intersection point of given lines will be center of the circle

$$\begin{gathered} 2x+4y=10-(2x-3y=-4)7y=14 \\ \Rightarrow y=2, x=1 \end{gathered}$$

So the equation of circle is ${\left(x-1\right)}^2+{\left(y-2\right)}^2=9$

$\Rightarrow x^2+y^2-2x-4y-4=0$