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

The Correct Option is B

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).