If the lines $2 x+3 y+1=0$ and $3 x-y-4=0$ lie along diameters of a circle of circumference $10 \pi$, then…
If the lines $2 x+3 y+1=0$ and $3 x-y-4=0$ lie along diameters of a circle of circumference $10 \pi$, then the equation of the circle is
$x^2+y^2-2 x+2 y-23=0$
$x^2+y^2-2 x-2 y-23=0$
$x^2+y^2+2 x+2 y-23=0$
$x^2+y^2+2 x-2 y-23=0$
Solution
Intersection of given lines is the centre of the circle i.e. $(1,-1)$ Circumference $=10 \pi \Rightarrow$ radius $r=5$
$\Rightarrow$ equation of circle is $x^2+y^2-2 x+2 y-23=0$