In a square $A B C D$ of side length $a$, suppose $A B$ and $A D$ are along the coordinate axes. Then, the…

In a square $A B C D$ of side length $a$, suppose $A B$ and $A D$ are along the coordinate axes. Then, the circle that circumscribes the square is
  1. $x^2+y^2+a(x+y)=0$
  2. $x^2+y^2-a(x+y)=0$
  3. $x^2+y^2+2 a(x+y)=0$
  4. $x^2+y^2-2 a(x+y)=0$

Solution

$B D$ is diameter.
$\therefore$ Circle is $ \begin{array}{rr} & (x-0)(x-a)+(y-a)(y-0)=0 \\ \Rightarrow & x^2+y^2-a(x+y)=0 \end{array} $

Asked in: AP EAMCET 2022 (07 Jul Shift 2)

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