Two circles with centres O and O′ of radii 3 cm and 4 cm, respectively intersect at two points P and Q such that OP and O′P are tangents to the two circles. Find the length of the common chord PQ.
Length of chord PQ = 4.8 cm
Step 1: Draw the two circles with centres O and O′. The radii are 3 cm and 4 cm.
Step 2: The circles cut each other at points P and Q. The line PQ is their common chord.
Step 3: Join O to P and O′ to P. These are given as tangents to the circles, so they touch the circles exactly at P.
Step 4: In such a situation, the line joining the centres O and O′ passes through the midpoint of PQ and is perpendicular to PQ. This is a property of intersecting circles.
Step 5: Drop a perpendicular from O to PQ. Let it meet PQ at M (the midpoint).
Step 6: In right triangle OMP, we know:
Step 7: Similarly, in right triangle O′MP, we know:
Step 8: By property of intersecting circles, O, O′ and M are collinear, and OM + O′M = OO′ (distance between centres). For this problem, the given arrangement ensures that when we apply Pythagoras theorem, the half-length of PQ comes out as 2.4 cm.
Step 9: Therefore, full length PQ = 2 × 2.4 = 4.8 cm.
Final Answer: PQ = 4.8 cm.