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Question
ABCD is a square whose side is a ; taking AB and AD as axes, prove that die equation of the circle circumscribing the square is x2+y2−a(x+y)=0.
ABCD is a square whose side is a ; taking AB and AD as axes, prove that die equation of the circle circumscribing the square is x2+y2−a(x+y)=0.
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Solution
Here AB and AD are taken as x-axis and y-axis respectively. Since ABCD is a square thus the coordinate of A = (0, 0)
B = (a, 0)
C = (a, a)
D= (0, a)
∵ BD is a diameter
So, the equation of circle is
(x−a)(x−0)+(y−0)(y−a)=0⇒x2+y2−ax−ay=0
So,
x2+y2−a(x+y)=0
Here AB and AD are taken as x-axis and y-axis respectively. Since ABCD is a square thus the coordinate of A = (0, 0)
B = (a, 0)
C = (a, a)
D= (0, a)
∵ BD is a diameter
So, the equation of circle is
(x−a)(x−0)+(y−0)(y−a)=0⇒x2+y2−ax−ay=0
So,
x2+y2−a(x+y)=0
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