Question

$ABCD$ is a square with side $a$. If $AB$ and $AD$ are taken as positive coordinate axes then equation of circle circumscribing 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 $x^2+y^2-ax-ay=0$

Let the radius of circle be a/2, since it is

half of the diameter which is a. The centre of

the circle is going to occur at (a/2, a/2), since that

is also the centre of square

Using $(x-h{)}^2+(y-k{)}^2=r^2$

where r is the radius and (h,k) is centre

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

$x^2-\frac{2ax}{2}+\frac{a^2}{4}+y^2-\frac{2ay}{2}+\frac{a^2}{4}=\frac{a^2}{2}$

$x^2+y^2+\frac{a^2}{2}-ax-ay=\frac{a^2}{2}$

$x^2+y^2=ax+ay.$

$x^2+y^2=a(x+y)\Rightarrow x^2+y^2-ax-ay=0.$