1. **Problem statement:** We have a circle with center $O$ and radius $OC=8$ units. The line segment $AC$ is tangent to the circle at point $C$, and $AC=15$ units. Points $A$, $B$, and $C$ lie on a line, and we want to find the length of $AB$. 2. **Key fact about tangents:** A tangent to a circle is perpendicular to the radius at the point of tangency. So, $OC \perp AC$. 3. **Setup:** Since $AC$ is tangent at $C$, and $O$ is the center, triangle $OBC$ is right-angled at $C$ if $B$ lies on the circle or inside it. But here, $B$ lies between $A$ and $C$ on the line, so $B$ is on segment $AC$. 4. **Using the right triangle $OBC$:** We know $OC=8$ (radius), and $AC=15$. Since $A$, $B$, and $C$ are collinear with $AC=15$, and $B$ lies between $A$ and $C$, let $AB=x$. Then $BC=15 - x$. 5. **Using the Pythagorean theorem in triangle $OBC$:** Since $OB$ is a segment from center $O$ to point $B$ inside the circle, and $OC$ is perpendicular to $AC$, triangle $OBC$ is right angled at $C$. So, $$OB^2 = OC^2 + BC^2 = 8^2 + (15 - x)^2 = 64 + (15 - x)^2.$$ 6. **Using triangle $ABO$:** Points $A$, $B$, and $O$ form a triangle with sides $AB=x$, $BO=OB$, and $AO=15 + 8 = 23$ (since $AO = AC + CO$ and $CO=8$). 7. **Applying the Law of Cosines in triangle $ABO$:** Since $AC$ is tangent and $OC$ is radius, angle $ACO$ is $90^\circ$. Thus, angle $AOB$ is $90^\circ$ because $O$, $B$, and $C$ form a right triangle with $C$ as the right angle. But since $B$ lies on $AC$, $A$, $B$, and $C$ are collinear, so angle $AOB$ is not $90^\circ$. Instead, we use the fact that $OB$ is the hypotenuse of triangle $OBC$. 8. **Using the fact that $OB$ is the hypotenuse of right triangle $OBC$ and $OB$ is also a side of triangle $ABO$:** We can write the length $AB=x$ and $AO=23$. 9. **Since $A$, $B$, and $C$ are collinear, and $AC=15$, $AB=x$, $BC=15-x$, and $OC=8$, the length $OB$ can be expressed as:** $$OB = \sqrt{64 + (15 - x)^2}.$$ 10. **Using the triangle $ABO$ and the fact that $O$, $B$, and $A$ form a triangle, we apply the triangle inequality or use the fact that $OB$ is the distance from $O$ to $B$ on the line segment $AC$. Since $B$ lies between $A$ and $C$, and $O$ is off the line $AC$, the distance $OB$ can be found by the Pythagorean theorem in triangle $OBC$ as above. We want to find $x=AB$ such that $OB$ is consistent with the geometry.** 11. **Since $O$ is at a distance $8$ from $C$ perpendicular to $AC$, and $A$ is $15$ units from $C$ along $AC$, the distance $AO=15+8=23$ is incorrect because $O$ is not on the line $AC$. Instead, $O$ is perpendicular to $AC$ at $C$, so the distance $AO$ is the hypotenuse of a right triangle with legs $AC=15$ and $OC=8$: $$AO = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17.$$ 12. **Now, triangle $ABO$ has sides $AB=x$, $AO=17$, and $OB=\sqrt{64 + (15 - x)^2}$. Since $A$, $B$, and $C$ are collinear, and $B$ lies between $A$ and $C$, the points $A$, $B$, and $O$ form a triangle where $AB$ and $OB$ are legs and $AO$ is the hypotenuse. So by the Pythagorean theorem:** $$AB^2 + OB^2 = AO^2.$$ 13. **Substitute $AB=x$, $OB=\sqrt{64 + (15 - x)^2}$, and $AO=17$:** $$x^2 + 64 + (15 - x)^2 = 17^2 = 289.$$ 14. **Expand and simplify:** $$x^2 + 64 + (225 - 30x + x^2) = 289,$$ $$x^2 + 64 + 225 - 30x + x^2 = 289,$$ $$2x^2 - 30x + 289 = 289,$$ $$2x^2 - 30x = 0,$$ $$2x(x - 15) = 0.$$ 15. **Solve for $x$:** $$x=0 \quad \text{or} \quad x=15.$$ 16. **Interpretation:** $x=0$ means $AB=0$ which is not possible since $B$ is between $A$ and $C$. $x=15$ means $AB=15$ units. **Final answer:** $$\boxed{15}$$ units. This means the length of $AB$ is 15 units.