If $3, 4$ are the radii and $5$ is the distance between the centres of two intersecting circles then the length of the common chord of the circles is :

Drop a perpendicular from the first circle's centre to the chord, forming a right triangle with hypotenuse 3 and base x. Using the Pythagoras theorem, x squared equals 3 squared minus h squared. Dropping a perpendicular from the second circle's centre forms another right triangle with hypotenuse 5, giving x squared equals 4 squared minus (5-h) squared. Equating the two yields 9 minus h squared equals 16 minus 25 plus 10h minus h squared, which solves to h equaling 1.8 or 9/5. Substituting h back gives x squared as 9 minus 81/25, which is 144/25, making x equal to 12/5. The total length of the common chord is 2x, which is 24/5.

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