A circle , with centre O , radius 6cm has chords AB and CD which intersects at x . Find OX , if AX = 3cm and XB = 2cm

OX = 5.4772 cm

Step-by-step 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

can you put the picture up?

step-by-step explanation:

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