In the figure given below, two circles with centres A and B touch externally. PM is a tangent to the circle with centre A and QN is a tangent to the circle with centre B. If PM = 15 cm, QN = 12 cm, PA = 17 cm and QB = 13 cm, then find the distance between the centres A and B of the circles.

Since radius and tangent at the point of contact of a circle are perpendicular to each other.

∴ ∠AMP = ∠BNQ = 90°.

In right angled triangle △AMP

    AP² = AM² + PM² (By pythagoras theorem)
⇒ AM² = AP² - PM²
⇒ AM² = 17² - 15²
⇒ AM² = 289 - 225
⇒ AM² = 64
⇒ AM = $\sqrt{64}$
⇒ AM = 8 cm.

Similarly in right angled triangle △BNQ

    BQ² = BN² + NQ² (By pythagoras theorem)
⇒ BN² = BQ² - NQ²
⇒ BN² = 13² - 12²
⇒ BN² = 169 - 144
⇒ BN² = 25
⇒ BN = $\sqrt{25}$
⇒ BN = 5 cm.

From figure the distance between A and B is equal to the sum of their radius = 8 + 5 = 13 cm.