+26367 Find the area of the square.
\(\text{Let the side of the square $=a$} \\ \text{Let the area of the square $=a^2$} \\ \text{Let the radius of the circle $=r$} \\ \text{Let the center of the circle $(x_c,y_c)=(-r,\ 0)$} \)
circle:
\(\begin{array}{|rcll|} \hline \mathbf{\text{circle:}} \\ \hline \mathbf{(x+r)^2+y^2} &=& \mathbf{r^2} \quad | \quad x=-\dfrac{a}{2} \\\\ \left(r-\dfrac{a}{2}\right)^2+y^2 &=& r^2 \\\\ r^2 -2\cdot \dfrac{ar}{2}+\dfrac{a^2}{4} +y^2 &=& r^2 \\\\ r^2 - ar +\dfrac{a^2}{4} +y^2 &=& r^2 \\ y^2 &=& ar -\dfrac{a^2}{4} & \boxed{2r = a-6 \\ r=\dfrac{a-6}{2}} \\ y^2 &=& a\left(\dfrac{a-6}{2}\right) -\dfrac{a^2}{4} \\\\ y^2 &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} & \boxed{4+y = \dfrac{a}{2}\\y=\dfrac{a}{2}-4\\y=\dfrac{a-8}{2} } \\ \left(\dfrac{a-8}{2}\right)^2 &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} \\\\ \dfrac{(a-8)^2}{4} &=& 2a\dfrac{(a-6)}{4} -\dfrac{a^2}{4} \\\\ (a-8)^2 &=& 2a(a-6) -a^2 \\ a^2-16a+64 &=& 2a^2-12a-a^2 \\ -16a+64 &=& -12a \\ 64 &=& 4a \quad | \quad :4 \\ 16 &=& a \\ \mathbf{a} &=& \mathbf{16} \\ \hline \end{array}\)
The area of the square is \(a^2 = 16^2 = \mathbf{256}\)
it's not 324 u² 
