X is a point 7cm from the center of circle O with radius 9 cm. AXB is a chord of the circle such that AX= 2 XB. Find the length of AB.

For the chord AB

k=mXB
2k=mAX

We will construct diameter CD such that X lies on CD.
Now CX=9+7=16 cm, XD=9-7=2 cm

Using the theorem of intersecting chords (AB and CD), we have
16*2=k(2k)
Solve for k.

To find the length of AB, we can use the Pythagorean Theorem and some algebra.

Let's label the center of the circle O and the point where the chord AXB intersects the circumference of the circle as C.

First, we find the length of OC, which is the radius of the circle. Given that the radius of the circle is 9 cm, OC = 9 cm.

Then, we can find the length of AC. Since X is a point 7 cm from the center of the circle, AC = OC - AX = 9 cm - 7 cm = 2 cm.

Now, let's focus on the chord AXB. We are told that AX = 2XB. Let's label the length of XB as y. Therefore, AX = 2y.

Since Y is the midpoint of AXB, the length of AY is also y. And since AY = AC - CY, we have y = AC - CY.

Substituting the values we know, we get y = 2 cm - CY.

Now, let's consider triangle ACY. Using the Pythagorean Theorem, we have AC^2 = AY^2 + CY^2.

Since we know AC = 2 cm and AY = y, we can substitute these values into the equation:

(2 cm)^2 = y^2 + CY^2.

Expanding this equation, we have 4 cm^2 = y^2 + CY^2.

Since we know that y = 2 cm - CY, let's substitute this into the equation:

4 cm^2 = (2 cm - CY)^2 + CY^2.

Expanding this equation, we have 4 cm^2 = 4 cm^2 - 4 cm * CY + CY^2 + CY^2.

Simplifying this equation, we get 4 cm^2 = 4 cm^2 - 4 cm * CY + 2 CY^2.

By canceling out the common terms, we get 0 = - 4 cm * CY + 2 CY^2.

Rearranging the terms, we have 2 CY^2 - 4 cm * CY = 0.

Factorizing the equation, we get 2 CY (CY - 2 cm) = 0.

Since we can't have a negative length for CY, we discard the CY - 2 cm = 0 solution and focus on CY = 0.

If CY = 0, then Y coincides with C, and thus A and B also coincide. In this case, the length of AB is zero.

Therefore, the length of AB is 0 cm.