Two circles of radius 15 cm and 37 cm intersect each other at the points A and B. If the length of common chord is 24 cm, what is the distance (in cm) between the centres of the circles?

44

AB = 24 cm

= AM = MB = 12 cm

In triangle AMO

\( { OM}^{2 } \) = \( {AO }^{ 2} \) - \( {AM }^{ 2} \)

= \( { OM}^{2 } \) = \( {37 }^{ 2} \) - \( {12 }^{ 2} \)

= \( { OM}^{2 } \) = 1369 - 144

= \( { OM}^{2 } \) = 1225

= OM = 35 cm

In triangle AMO'

\( { O'M}^{2 } \) = \( {AO' }^{ 2} \) - \( {AM }^{ 2} \)

=\( { O'M}^{2 } \) = \( {15 }^{ 2} \) - \( {12 }^{ 2} \)

= \( { O'M}^{2 } \) =225 - 144

= \( { O'M}^{2 } \) = 81

= O'M = 9 cm

Now

O'O = OM + O'M  = (35 + 9)

= 44 cm

Therefore, the distance between the center of the circles is 44 cm.