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Abcd is a Square Whose Side is A; Taking Ab and Ad as Axes, Prove that the Equation of the Circle Circumscribing the Square is X2 + Y2 − a (X + Y) = 0.

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Question

ABCD is a square whose side is a; taking AB and AD as axes, prove that the equation of the circle circumscribing the square is x2 + y2 − a (x + y) = 0.

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Solution

Given:
ABCD is a square with side a units.
Let AB and AD represent the x-axis and the y-axis, respectively.
Thus, the coordinates of B and D are (a, 0) and (0, a), respectively.
The end points of the diameter of the circle circumscribing the square are B and D.
Thus, equation of the circle circumscribing the square is

\[\left( x - a \right)\left( x - 0 \right) + \left( y - 0 \right)\left( y - a \right) = 0\]  or  \[x^2 + y^2 - a\left( x + y \right) = 0\]
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Advanced Concept of Circle - Standard Equation of a Circle
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Chapter 24: The circle - Exercise 24.3 [Page 37]

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R.D. Sharma Mathematics [English] Class 11
Chapter 24 The circle
Exercise 24.3 | Q 9 | Page 37

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