If the lines $3 x-4 y-7=0$ and $2 x-3 y-5=0$ are two diameters of a circle of area $49 \pi$ square units,…
If the lines $3 x-4 y-7=0$ and $2 x-3 y-5=0$ are two diameters of a circle of area $49 \pi$ square units, the equation of the circle is
$x^2+y^2+2 x-2 y-47=0$
$x^2+y^2+2 x-2 y-62=0$
$x^2+y^2-2 x+2 y-62=0$
$x^2+y^2-2 x+2 y-47=0$
Solution
Point of intersection of $3 x-4 y-7=0$ and $2 x-3 y-5=0$ is $(1,-1)$, which is the centre of the circle and radius $=7$.
$\therefore$ Equation is $(x-1)^2+(y+1)^2=49 \Rightarrow x^2+y^2-2 x+2 y-47=0$