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?
Answer & explanation
Correct answer: option 1
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.