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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
Given lines
L1:x+y6=0L2:2x+y4=0L3:x+2y5=0
Consider the equation
L1L2+λL2L3+μL3L1=0
(x+y6)(2x+y4) +λ(2x+y4)(x+2y5) +μ(x+2y5)(x+y6)=0 (i)
Where λ and μ are constants.

This is a circle, when
coefficient of x2= coefficient of y2
2+2λ+μ=1+2λ+2μμ=1

Coefficient of xy=0
3+5λ+3μ=0λ=65

On substituting the values of λ and μ in equation (i), we get
L1L265L2L3+L3L1=05L1[L2+L3]=6L2L35(x+y6)3(x+y3)=6(2x+y4)(x+2y5)5[x2+y2+2xy9x9y+18] =2[2x2+2y2+5xy14x13y+20]x2+y217x19y+50=0


Alternate Solution:
Let the equation of side be
AB:x+y=6BC:2x+y=4AC:x+2y=5

Finding the point of intersection, we get the coordinates of the vertices as
A=(7,1), B=(2,8), C=(1,2)

Let the equation of the circumcircle of ABC be
x2+y2+2gx+2fy+c=0 (ii)

It passes through A(7,1),B(2,8) and C(1,2), so
50+14g2f+c=0684g+16f+c=05+2g+4f+6c=0
On solving these equations, we get
g=172,f=192, c=50

Hence, the equation of the circumcircle is
x2+y217x19y+50=0

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