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Question

The equation of the circle circumscribing the triangle formed by the lines x+y=6,2x+y=4 and x+2y=5 is

A
x2+y217x19y+50=0
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B
x2+y2+17x+19y+50=0
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C
x2+y219x17y+50=0
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D
x2+y2+19x17y+50=0
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Solution

The correct option is A x2+y217x19y+50=0
Let the equation of side AB,BC and CA of ABC are respectively
x+y=6,
2x+y=4
and x+2y=5

The coordinates of A,B and C are (7,1),(2,8) and (1,2) respectively.
Let the equation of the circumcircle of ABC be
x2+y2+2gx+2fy+c=0 (i)
It passes through A(7,1),B(2,8) and C(1,2)
Therefore,
50+14g2f+c=0
684g+16f+c=0
and 5+2g+4f+6c=0
On solving these equations, we get
g=172,f=192 and c=50
On substituting the value of g,f and c in Equation (i), the equation of the required circumcircle is
x2+y217x19y+50=0

Aliternative
Consider the equation
(x+y6)(2x+y4)+λ(2x+y4)(x+2y5)+μ(x+2y5)(x+y6)=0
where λ and μ are constants.
This equation represents a curve passing through the vertices of the triangle formed by the given lines.
We have to determine the values of λ and μ so that the curve given in Equation (i) is a circle.
The curve given in Equation (i) will be a circle, if
coefficient of x2 = coefficient of y2
and coefficient of xy=0
2+2λ+μ=1+2λ+2μ3+5λ+3μ=0
μ=1 and 3μ+5λ+3=0
μ=1 and λ=65
On substituting the values of λ and μ in Eq. (i), we get
(x+y6)(2x+y4)(65)(2x+y4)(x+2y5)+(x+2y5)(x+y6)=0
x2+y217x19y+50=0

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