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Question

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

A
x2+y217x19y+50=0
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B
x2+y2+17x+19x+50=0
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C
x2+y219x17x+50=0
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D
x2+y219x17x50=0
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Solution

The correct option is A x2+y217x19y+50=0
Let the sides of ABC be
AB:x+y=6 (i)BC:2x+y=4 (ii)AC:x+2y=5 (iii)

The coordinates of A,B and C are
A=(7,1)B=(2,8)C=(1,2)

Let the equation of the circumcircle of ABC be
x2+y2+2gx+2fy+c=0
It passes through A(7,1),B(2,8) and C(1,2), so
50+14g2f+c=0 (iv)684g+16f+c=0 (v)5+2g+4f+c=0 (vi)

On solving these equations, we get
g=172,f=192 and c=50

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

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