Question

Find the equation to the circle circumscribing the quadrilateral formed by the straight lines $2x+3y=2,3x-2y=4,x+2y=3$, and $2x-y=3$.

Answer 1

The equation of circle circumscribing a quadrilateral whose sides in order are represented by the lines $L_1=0,L_2=0,l_3=0,l_4=0$ is $L_1{L}_3+\lambda L_2{L}_4=0$ provided coefficient of$X^2$ equals coefficient of $y^2$ and coefficient of $xy=0$

So, the equation of circle becomes

$(2x+3y-2)(x+2y-3)+\lambda (3x-2y-4)(2x-y-3)=0$

Equating the coefficient of $x^2=y^2$, we get

$2+\lambda =6+2\lambda$

$\lambda =1$

Putting $\lambda =1$ in the equation of circle, we get

$8x^2+8y^2-25x-3y+18=0$ is the final equation of circle.