Question

ABCD is a square with side "a: If AB and AD are taken as positive coordinate axes then equation of circle circunscribing the square is

• A. $x^2+y^2-ax-ay=0$
• B. $x^2+y^2+ax+ay=0$
• C. $x^2+y^2-ax+ay=0$
• D. $x^2+y^2+ax-ay=0$

Answer 1

The correct option is A $x^2+y^2-ax-ay=0$

$Centreofsuchcircle=\left(\right(\frac{a}{2}),(\frac{a}{2}\left)\right)andradius=\sqrt{(\frac{a}{2}{)}^2+(\frac{a}{2}{)}^2}=\sqrt{\left(\frac{a^2}{4}\right)+\left(\frac{a^2}{4}\right)}=\left(\frac{a}{\sqrt{2}}\right)\therefore e{q}^{n}ofcircle{(x-(\frac{a}{2}\left)\right)}^2+{(y-(\frac{a}{2}\left)\right)}^2={\left(\right(\frac{a}{\sqrt{2}}\left)\right)}^2{x}^2-ax+\left(\frac{a^2}{4}\right)+y^2-ay+\left(\frac{a^2}{4}\right)=\left(\frac{a^2}{2}\right)x^2+y^2-ax-ay=0$