Let ABCD and AEFG be squares of side 4 and 2 | vedclass

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(B) Let the square $ABCD$ have vertices $A(0,0)$,$B(4,0)$,$C(4,4)$,and $D(0,4)$.The diagonal $AC$ lies on the line $y=x$.Since $AEFG$ is a square of s

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$r^2-8r+8=0$

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(B) Let the square $ABCD$ have vertices $A(0,0)$,$B(4,0)$,$C(4,4)$,and $D(0,4)$.The diagonal $AC$ lies on the line $y=x$.Since $AEFG$ is a square of side $2$ with $E$ on $AB$ and $F$ on $AC$,the coordinates are $A(0,0)$,$E(2,0)$,$G(0,2)$,and $F(2,2)$.The circle touches the lines $BC$ $(x=4)$ and $CD$ $(y=4)$.Let the center of the circle be $O(h,k)$. Since it touches $x=4$ and $y=4$ with radius $r$,the center is $O(4-r, 4-r)$.The circle passes through $F(2,2)$,so the distance $OF=r$.Using the distance formula: $(4-r-2)^2 + (4-r-2)^2 = r^2$.$(2-r)^2 + (2-r)^2 = r^2$.$2(4 - 4r + r^2) = r^2$.$8 - 8r + 2r^2 = r^2$.$r^2 - 8r + 8 = 0$.

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:

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