The length of common chord of two intersecting circles is 30 cm. If the diameters of these two circles be 50 cm and 34 cm, calculate the distance between their centers.

Let O be the center of the circle with diameter 50 cm and O' be the center of the circle with diameter 34 cm.

Let AB be the common chord.

Radius = $\dfrac{\text{Diameter}}{2}$

Radius of circle with center O = $\dfrac{50}{2}$ = 25 cm,

Radius of circle with center O' = $\dfrac{34}{2}$ = 17 cm.

We know that,

Perpendicular from center to chord, bisects the chord.

∴ AC = BC = $\dfrac{AB}{2} = \dfrac{30}{2}$ = 15 cm.

In right-angled triangle AOC,

By pythagoras theorem,

⇒ AO² = AC² + OC²

⇒ 25² = 15² + OC²

⇒ 625 = 225 + OC²

⇒ OC² = 625 - 225

⇒ OC² = 400

⇒ OC = $\sqrt{400}$ = 20 cm.

In right-angled triangle AO'C,

By pythagoras theorem,

⇒ AO'² = AC² + O'C²

⇒ 17² = 15² + O'C²

⇒ 289 = 225 + O'C²

⇒ O'C² = 289 - 225

⇒ O'C² = 64

⇒ O'C = $\sqrt{64}$ = 8 cm.

From figure,

OO' = OC + O'C = 20 + 8 = 28 cm.

Hence, distance between centers = 28 cm.