## Circles Within Square

The two circles in the figure are congruent and have their centers on the line $\overleftrightarrow{AC}$. They are tangent to each other and to the square $ABCD$. The radius of the circles is $1$. Calculate the area of the square $ABCD$.

Source: NCTM Mathematics Teacher, December 2005

SOLUTION

$\overline{CE}$ is tangent to the left circle at point $E$ and the secant segment $\overline{CO}$ intersects the circle at points $F$ and $O$
$CE^2=CF\times CO$
$1^2=CF\times CO$
$1=CF(CF+FO)$
$1=CF(CF+2)$
$1=CF^2+2CF$
$0=CF^2+2CF-1$
Using the quadratic formula
$CF=\textrm{-}1+\sqrt 2$
$CO=CF+FO$
$=\textrm{-}1+\sqrt 2+2$
$=1+\sqrt 2$
Triangle $COB$ is a $45^\circ\!\textrm{-}45^\circ\!\textrm{-}90^\circ$ triangle with side $CO=1+\sqrt 2$
$CB=CO\sqrt 2$
$=(1+\sqrt 2)\sqrt 2$
$=\sqrt 2+2$
Area of square $ABCD$
$CB^2=(\sqrt 2+2)^2$
$=2+4\sqrt 2+4$
$=6+4\sqrt 2$

Answer: $6+4\sqrt 2$ square units

Alternative solution

Triangle $GHI$ is a $45^\circ\!\textrm{-}45^\circ\!\textrm{-}90^\circ$ triangle with hypotenuse $GI=2$
$GH\sqrt 2=GI$
$GH\sqrt 2=2$
$GH=\sqrt 2$
$CB=CJ+JK+KB$
$=1+\sqrt 2+1$
$=2+\sqrt 2$
Area of square $ABCD$
$CB^2=(2+\sqrt 2)^2$
$=4+4\sqrt 2+2$
$=6+4\sqrt 2$