ABCD is a square of area $4$ sq. units. Taking AB and AD as the axes of the coordinates, the equation of the circle which touches the sides of the square, is

x^{2} + y^{2} - 2 x - 2 y - 1 = 0

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x^{2} + y^{2} - 2 x - 2 y + 1 = 0

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x^{2} + y^{2} - 4 x - 4 y + 1 = 0

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x^{2} + y^{2} - 4 x - 6 y - 6 = 0

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Solution

The correct option is B $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$
Area of $□ A B C D = 4$ ....Given
$∴ A B = 2$ .
$E q^{n}$ of the circle having centre $(1, 1)$ and radius $1$ units and touching all sides of the square $A B C D$ is given by
$(x - 1)^{2} + (y - 1)^{2} = 1$
$\Rightarrow x^{2} + y^{2} - 2 x - 2 y + 1 = 0$

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ABCD is a square of area 4 sq. units. Taking AB and AD as the axes of the coordinates, the equation of the circle which touches the sides of the square, is

The correct option is B x2+y2−2x−2y+1=0Area of □ABCD=4 ....Given∴AB=2.Eqn of the circle having centre (1,1) and radius 1 units and touching all sides of the square ABCD is given by(x−1)2+(y−1)2=1⇒x2+y2−2x−2y+1=0

ABCD is a square of area $4$ sq. units. Taking AB and AD as the axes of the coordinates, the equation of the circle which touches the sides of the square, is