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Points B and D are points of tangency. Find the two possible lengths of AB.

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If B and D are tangents then line BC is perpendicular to AB. Similarly CD is perpendicular to AD.

Then triangle ABC and ADC become right triangles.

For both triangles AC is the hypotenuse. BC and CD are radius of the circle.

By Pythagoras theorem;

`AC^2 = AB^2+BC^2`

`AC^2 = AD^2+CD^2`

`AD^2+CD^2 = AB^2+BC^2`

But `BC = CD` (radius of circle)

`AD^2 = AB^2`

`AD = AB`

`x+10 = x^2+x-6`

`x^2 = 16`

`x = +-4`

When x = 4

`AB = x^2+x-6 = 16+4-6 = 14`

When x = -4

`AB = x^2+x-6 = (-4)^2-4-6 = 6`

*So the possible length of AB is 14 and 6.*

**Sources:**

A tangents to a circle from an exterior point are equal in length.

So

AB=AD

AB=x^2+x-6

AD=x+10

Thus

`x^2+x-6=x+10`

`x^2=10+6`

`x^2=16`

`x=+-4`

`If `

`x=4`

`AB=16+4-6=14`

`AD=4+10=14`

If x=-4

AB=16-4-6=6

AD=-4+10=6

Thus possible length of two tangents are 14 or 6

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