ABCD is a square, the length of whose side is a. Taking AB and AD as the coordinate axes, the equation of the

circle passing through the vertices of the square, is

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Solution

The correct option is B

According to the figure A(0, 0), B(a, 0) and D(0, a)

and centre is $(\frac{a}{2}, \frac{a}{2})$ .

Therefore, the equation of circle

${(x - \frac{a}{2})}^{2} + {(y - \frac{a}{2})}^{2} = \frac{a^{2}}{2}$

$\Rightarrow x^{2} + y^{2} - a x - a y = 0.$

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ABCD is a square, the length of whose side is a. Taking AB and AD as the coordinate axes, the equation of the circle passing through the vertices of the square, is

The correct option is B According to the figure A(0, 0), B(a, 0) and D(0, a) and centre is (a2,a2). Therefore, the equation of circle (x−a2)2+(y−a2)2=a22 ⇒x2+y2−ax−ay=0.

ABCD is a square, the length of whose side is a. Taking AB and AD as the coordinate axes, the equation of the

circle passing through the vertices of the square, is

The correct option is B

According to the figure A(0, 0), B(a, 0) and D(0, a)

and centre is $(\frac{a}{2}, \frac{a}{2})$ .

Therefore, the equation of circle

${(x - \frac{a}{2})}^{2} + {(y - \frac{a}{2})}^{2} = \frac{a^{2}}{2}$

$\Rightarrow x^{2} + y^{2} - a x - a y = 0.$