Two circles of radii 4 cm and 3 cm intersect at two points and the distance between their centres is 5 cm. Find the length of the common chord.
Consider the figure:
Given: AC=3 cm, BC=4 cm and AB=5 cm
Since, AB2=BC2+AC2, △ABC is a right-angled triangle, right angled at C.
Let AD be x cm. Then BD=5−x cm
Applying Pythagoras theorem to △ADC,
⇒CD2=9−x2 ... (i)
Applying Pythagoras theorem to △BDC,
⇒CD2=16−(5−x)2 ... (ii)
Equating equation (i) and (ii), we get,
We know that the perpendicular from a centre to a chord bisects the chord.
So, length of the common chord =2×CD=2×2.4=4.8 cm