In the given figure, $PT$ is a tangent to the circle centered at $O$. $OC$ is perpendicular to chord $AB$. Prove that $PA \cdot PB = PC^2 - AC^2$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
Given: OC ⊥ AB, so C is the midpoint of AB (perpendicular from centre bisects chord).
Thus, $AC = BC$.
Since P, C, O are collinear and AB is a chord, using the property of intersecting lines:
$$PA \cdot PB = (PC - AC)(PC + AC) = PC^2 - AC^2$$
(Since $PA = PC - AC$ and $PB = PC + BC = PC + AC$)
Hence, $PA \cdot PB = PC^2 - AC^2$. [Proved]
Explanation
- Key fact 1: OC ⊥ AB ⟹ C is the midpoint of AB, so AC = BC.
- Key fact 2: Since P, C, B, A are collinear (P is external, C is foot of perpendicular on chord), we get PA = PC − AC and PB = PC + AC (C lies between P and B, A lies between C and B is not the case — actually A and B are on either side of C, and P is beyond C externally).
- The result follows from the difference-of-squares identity: $(PC - AC)(PC + AC) = PC^2 - AC^2$.
- Examiners want you to clearly state why AC = BC and then substitute PA and PB in terms of PC and AC.