Question

PQ is a transverse common tangent to the circles with centers A and B touching them at P and Q respectively: Prove that \(\frac{AP}{BQ}\) = \(\frac{AO}{BO}\) where O is the point of intersection of the common tangent and the line joining the centers.

Answer 1

PQ is the tangent to the circle with center A at point P and to the circle with center B at point Q. 

∴ ∠ APQ = 90° (radius ⊥ tangent at the point of contact) 

Similarly ∠ BQP = 90° 

∴ In ∆ APO and BQO 

∠ APO = ∠ BQO (each = 90°) 

∠ AOP = ∠ BOQ (vert. opp. ∠s) 

∴ ∆ APO ~ ∆ BQO (AA similarity) 

⇒ \(\frac{AP}{BQ} = \frac{AO}{BO}.\)