Answer:
OX = 5.4772 cm
Explanation:
Please check image attached for the drawing of the circle with the chords and points.
From the theorem of intersecting chords, we have:
AE * EB = EF * EG
With AE = AX - EX, EB = BX + EX, EF = 6 - EO and EG = 6 + EO, we have:
(AX - EX) * (BX + EX) = (6 - EO) * (6 + EO)
(3 - EX) * (2 + EX) = 36 - EO^2
6 + EX - EX^2 = 36 - EO^2
EO^2 - EX^2 + EX = 30 (eq1)
From the triangle AEO, we have:
AE^2 + EO^2 = OA^2
(AX - EX)^2 + EO^2 = 6^2
(3 - EX)^2 + EO^2 = 36
9 - 6*EX + EX^2 + EO^2 = 36
EX^2 - 6*EX + EO^2 = 27 (eq2)
If we do (eq1) - (eq2), we have:
-2*EX^2 + 7*EX = 3
2*EX^2 - 7*EX + 3 = 0
Solving this quadratic equation, we have EX = 3 cm or EX = 0.5 cm
EX cannot be 3 cm, because AE would be 0 cm, so EX = 0.5 cm
Calculating EO, we have:
EO^2 - 0.5^2 + 0.5 = 30
EO^2 = 29.75
EO = 5.4544 cm
Now, using Pythagoras in the triangle EOX, we have:
EO^2 + EX^2 = OX^2
29.75 + 0.25 = OX^2
OX^2 = 30
OX = 5.4772 cm