The equation of a circle which passes through (1,2) and (2,1) and whose radius is 1 units is

x²+y²-4x-4y+7=0

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x²+y²-2x-4y+4=0

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x²+y²-4x-2x+4=0

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x²+y²-x-y+1=0

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Solution

The correct option is B
x²+y²-2x-4y+4=0

$x^{2} + y^{2} + 2 g x + 2 f y + c = 0$
It passess through (1,2),(2,1) and has a radius of 1,2
So
$5 + 2 g + 4 F + C = 0 \dots 1$
$5 + 4 g + 2 F + C = 0 \dots 2$
Solving 1 and 2, we get
g=F
$\sqrt{g^{2} + F^{2} - C} = 1$
$C = 2 g^{2} - 1 \dots 3$
From the three equations we get
g=-2 or -1,F=-2 or -1, c=7 or 1.
So equation of the circle is
$x^{2} + y^{2} - 4 x - 4 y + 7 = 0$ or
$x^{2} + y^{2} - 2 x - 2 y + 1 = 0$

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