Question

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 :

• A. r = 1
• B. $\ r^{2} - 8r + 8 = 0$
• C. $\ 2r^{2} - 4r + 1 = 0$
• D. $\ 2r^{2} - 8r + 7 = 0$

Answer 1

Answer: (B)

$\ r^{2} - 8r + 8 = 0$

Answer 2

Consider the square ABCD with vertices at:

$A(0, 0), \, B(4, 0), \, C(4, 4), \, D(0, 4).$

The point $E$ lies on the line segment $AB$ with coordinates:

$E(2, 0).$

The point $F$ lies on the diagonal $AC$ at:

$F(2, 2).$

Geometry of the Circle Let the radius of the circle be $r$, and let $O$ be the center of the circle. The circle passes through $F(2, 2)$ and touches the line segments $BC$ (at $x = 4$) and $CD$ (at $y = 4$).

To find the equation of the circle, we use the condition that the distance between the center $O$ and the lines $BC$ and $CD$ must be equal to the radius $r$.

Distance Calculation From the geometry:

$OF^2 = r^2.$

Using the distance formula, we find:

$(2 - r)^2 + (2 - r)^2 = r^2.$

Simplifying:

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

Therefore, the correct answer is Option (2).