AC is a transverse common tangent to two circle with centres P and Q and radii 6 cm and 3 cm at the point A and C respectively. If AC cuts PQ at the point B and AB = 8 cm, then the length of PQ is:

Solution (By Examveda Team)

According to the question,
⇒ AP = 6 cm   (Radius₁)
⇒ QC = 3 cm   (Radius₂)
As we know, any line drawn from centre to the tangent is perpendicular.
⇒ So, ∠PAB = ∠QCB = 90°
⇒ ∠APB = ∠CQB = θ   (same alternative angle)
⇒ So, ΔAPB ∽ ΔCQB
$$\eqalign{ & \Rightarrow \frac{{{\text{AP}}}}{{{\text{CQ}}}} = \frac{{{\text{AB}}}}{{{\text{CB}}}} \cr & \Rightarrow \frac{6}{3} = \frac{8}{{{\text{CB}}}} \cr} $$
⇒ CB = 4 cm
⇒ In right angled triangle ΔPAB
⇒ (PB)² = (PA)² + (AB)²
⇒ (PB)² = 6² + 8²
⇒ PB = 10 cm
⇒ Again, in right angled triangle ΔCQB
⇒ (BQ)² = (BC)² + (CQ)²
⇒ (BQ)² = 3² + 4²
⇒ BQ = 5 cm
⇒ Therefore PQ = PB + BQ
⇒ PQ = 10 + 5
⇒ PQ = 15 cm