Length Ab

1. **State the problem:** We have a circle centered at point O. Line segment \(\overleftrightarrow{AC}\) is tangent at point C. We know \(AC = 15\) units and \(OC = 8\) units (perpendicular to AC). Point B lies on segment AO. We need to find the length \(AB\). 2. **Analyze the geometry:** Since \(OC\) is perpendicular to tangent \(AC\) at point C, \(OC\) is the radius of the circle, so \(OC = 8\). 3. **Locate points:** On line segment \(AO\), point B lies such that \(AB\) and \(BC\) form a right triangle \(ABC\). 4. **Find length \(AO\):** Since \(AC = 15\) and \(OC = 8\), and \(O\) lies somewhere along the vertical line, we consider the right triangle \(AOC\) with right angle at C. Using Pythagoras theorem: $$AO = \sqrt{AC^2 + OC^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = 17.$$ 5. **Interpret point B:** Since B lies on segment AO where triangle ABC is right angled at B, triangle ABC has right angle at B. 6. **Apply the right angle property:** Triangle ABC right angled at B implies: $$AB^2 + BC^2 = AC^2.$$ 7. **Express lengths:** \(AB = AB\) (unknown), \(BC = OC - OB\) since B is on AO, and O lies between A and B. 8. **Coordinates approach:** Assign coordinates: Let A at 0 on x-axis, C at 15, O at some point such that OC = 8 vertical. Since AC is horizontal and length 15, and OC is vertical 8, O is at (15,8). 9. Thus: A = (0,0), C = (15,0), O = (15,8) B lies on AO, which is from A(0,0) to O(15,8). 10. Parameterize B along AO: $$B = (15t, 8t), 0 \leq t \leq 1.$$ 11. Calculate lengths: $$AB = \text{distance}(A,B) = \sqrt{(15t - 0)^2 + (8t - 0)^2} = t \sqrt{225 + 64} = 17t.$$ $$BC = \text{distance}(B,C) = \sqrt{(15 - 15t)^2 + (0 - 8t)^2} = \sqrt{(15(1 - t))^2 + (8t)^2} = \sqrt{225(1 - t)^2 + 64 t^2}.$$ 12. Use right angle at B: $$AB^2 + BC^2 = AC^2 = 15^2 = 225.$$ Substitute: $$(17 t)^2 + 225 (1 - t)^2 + 64 t^2 = 225$$ $$289 t^2 + 225 (1 - 2 t + t^2) + 64 t^2 = 225$$ 13. Expand and simplify: $$289 t^2 + 225 - 450 t + 225 t^2 + 64 t^2 = 225$$ $$ (289 + 225 + 64) t^2 - 450 t + 225 = 225$$ $$578 t^2 - 450 t + 225 = 225$$ 14. Subtract 225 both sides: $$578 t^2 - 450 t = 0$$ 15. Factor: $$t (578 t - 450) = 0$$ 16. Solutions for t: $$t = 0 \quad \text{or} \quad t = \frac{450}{578} = \frac{225}{289}.$$ 17. \(t=0\) corresponds to point A, so discard. 18. Therefore, $$t = \frac{225}{289} \approx 0.778.$$ 19. Find \(AB\): $$AB = 17 t = 17 \times \frac{225}{289} = \frac{3825}{289} \approx 13.24.$$ **Final answer:** \(AB \approx 13.24\) units.