Question

ABCD is a square of unit area. A circle is tangent to two sides of ABCD and passes through exactly one of its vertices. The radius of the circle is

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

Answer 1

Answer: (A)

$2-\sqrt{2}$

Let the radius of the circle is $(r,r)$

$OE=DE=(1-r)$

Applying Pythagoras theorem in $\Delta OED$

$\Rightarrow (1-r{)}^2+(1-r{)}^2=r^2$

$\Rightarrow 2(1-r{)}^2=r^2$

$\Rightarrow \sqrt{2}(1-r)=r$

$\Rightarrow r(\sqrt{2}+1)=\sqrt{2}$

$\Rightarrow r=\frac{\sqrt{2}}{\sqrt{2}+1}=\sqrt{2}(\sqrt{2}-1)=2-\sqrt{2}$