Let $ABCD$ and $AEFG$ be squares of side 4 and 2 units, respectively. The point $E$ is on the line segment $AB$ and the point $F$ is on the diagonal $AC$. Then the radius $r$ of the circle passing through the point $F$ and touching the line segments $BC$ and $CD$ satisfies:
Step-by-Step Solution
Key Concept: Place $A=(0,0)$, $B=(4,0)$, $C=(4,4)$, $D=(0,4)$. $E=(2,0)$ on $AB$, $F=(2,2)$ on diagonal $AC$. Circle touches $BC$ ($x=4$) and $CD$ ($y=4$): centre $(4-r,4-r)$. Distance from centre to $F(2,2)$ equals $r$.
Centre $(4-r,4-r)$, $F=(2,2)$: $2(2-r)^2=r^2\Rightarrow r^2-8r+8=0$.
Correct Answer: 4