Question

Let a circle is touching exactly two sides of a square $ABCD$ and passes through exactly one of its vertices. If area of the square $ABCD$ is $1$ sq. units, then the radius of the circle (in units) is

• A. $2-\sqrt{2}$
• B. $\sqrt{2}-1$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $2-\sqrt{2}$

Let $A=(0,0)$, so
$B=(0,1),C=(1,1),D=(1,0)$


The centre of the circle is $O=(r,r)$, so the equation of the circle is
$(x-r{)}^2+(y-r{)}^2=r^2$
It passes through $(1,1)$, so
$(1-r{)}^2+(1-r{)}^2=r^2\Rightarrow 2(1-r{)}^2=r^2\Rightarrow 2+2r^2-4r=r^2\Rightarrow r^2-4r+2=0\Rightarrow r=\frac{4\pm \sqrt{8}}{2}\therefore r=2-\sqrt{2}(\because r<1)$