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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 :))
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