In a triangle the angle made the tangent at any point with the radial line at that point is equal to 90 degrees.

From the figure, the sides of the triangle CAB are CA = 5 + 8 = 13, AB = 12 and BC = 5. 12^2 + 5^2 = 144 + 25 = 169 = 13^2. As the sum of the square of the length of sides CB and AB is equal to the square of the length of side AC, the triangle CBA is a right triangle with hypotenuse AC. As angle B is a right triangle, the sides AB and CB are perpendicular to each other. From the property of all tangents to a circle given in the beginning it can be concluded that AB is a tangent of the circle.

**The line AB is a tangent of the circle with center C.**

Produce AC ,meets the circle at D.

AD=DC+CA

=5+13

=18

If I call point E ,where AD meets circle.

Now we have two segments AD and AE.

Apply now secant-tangent theorem ,to test tangency

Let AB is tangent to circle.

Then

AD x AE= (AB)^2

18 x 8 =(12)^2

which is true ,hence our assumption that AB is tangent is true.

So you can say AB is tangent to the circle.