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Question

Find the equation of the circle with passing through three points (0,2),(3,0) and (3,2).

A
x2+y23x2y=0
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B
x2+y2+3x2y=0
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C
x2+y23x+2y=0
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D
x2+y2+3x+2y=0
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Solution

The correct option is B x2+y23x2y=0
Three points on the circle are:A(0,2),B(3,0),C(3,2)
Let P(x,y) be its centre, then |PA|=|PB|=|PC|
(x0)2+(y2)2=(x3)2+(y0)2=(x3)2+(y2)2
x2+y2+44y=x2+96x+y2=x2+96x+y2+44y
44y=96x=96x+44y
Taking first and third parts of the above equation,
44y=9+46x4y6x=9x=96=32
Taking second and third parts of the above equation,
96x=9+46x4y4y=4y=1P(32,1)
So, radius r=PA=(32)2+(12)2=94+1=52
Thus, equation of the circle ;_(x32)2+(y1)2=(52)2
x2+y23x2y+154+94=0x2+y23x2y=0

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