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 :
- r = 1
- 2r2 - 4r + 1 = 0
- 2r2 - 8r + 7 = 0
- r2 - 8r + 8 = 0
Answer
(4) r2 - 8r + 8 = 0
Solution
Take A as origin: B(4, 0), C(4, 4), D(0, 4), E(2, 0), F(2, 2)
[figure]
the circle touches BC (x = 4) and CD (y = 4) centre O(4 - r, 4 - r)
OF = r (2 - r)2 + (2 - r)2 = r2
2r2 - 8r + 8 = r2 r2 - 8r + 8 = 0
Answer: (4)