Question

$ABCD$ is a square whose side is $a$; taking $AB$ and $AD$ as axes, prove that the equation to the circle circumscribing the square is $x^2+y^2=a(x+y)$.

Answer 1

As the square is aligned along the axes so the centre of the circle will be the mid point of $AC$

$\Rightarrow$ Centre $O=(\frac{a}{2},\frac{a}{2})$

$AC=\sqrt{{(a-0)}^2+{(a-0)}^2}=\sqrt{2}a$

Radius of circle $=r=\frac{AC}{2}=\frac{\sqrt{2}a}{2}=\frac{a}{\sqrt{2}}$

So, the equation of circle with centre $O$ and radius $r$ is

${(x-\frac{a}{2})}^2+{(y-\frac{a}{2})}^2={\left(\frac{a}{\sqrt{2}}\right)}^2{x}^2+\frac{a^2}{4}-ax{+y}^2+\frac{a^2}{4}-ay=\frac{a^2}{2}{x}^2{+y}^2-ax-ay=0x^2{+y}^2=a(x+y)$

Hence proved.