Find the equation of the circumcircle of a triangle whose sides are given by the equations 2x + y - 3= 0, x - 2y + 1 = 0 and 3x - y - 7 = 0

Equation of a circle circumscribing a triangle whose sides are given by L1 = 0, L2 = 0,  L3 = 0.

This equation is given by L1L2  + λL2L3 + µL3 L1 = 0, provided coefficeint of xy = 0 and coefficient of x² = coefficient of y².

The particular value of the parameter λ and µ gives a unique circle. 

so, eqn here becomes (2x + y - 3)(x - 2y + 1) + λ(x - 2y + 1)(3x - y - 7) + µ(3x - y - 7)(2x + y - 3)= 0

or 2x^2-3xy-x-2y^2+7y-3+3mx^2-7mxy-4mx+2my^2+13my-7m+nxy+6nx^2-23nx-ny^2-4ny+21n= 0

so 2+3m+6n= – 2+2m – n

and – 3 – 7m+n= 0

so that m= n= – ½ 

so circle eqn becomes x^2 – 5x – y+y^2+4=0

KINDLY APPROVE :))