Q) Two circles with centres O and O’ of radii 6 cm and 8 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.
Ans:
Since OP is O’P
OO’2 = OP2 + O’P2 = 62 + 82 = 100
OO’ = 10 cm
Let’s consider OA = x, therefore AO’ = 10 – x
Now in Δ POA, AP2 = OP2 – OA2 = 36 – x2 ……………….. (i)
Similarly, Now in Δ PO’A, AP2 = O’P2 – O’A2 = 64 – (10 – x)2 …………. (ii)
From equations (i) and (ii), we get:
36 – x2 = 64 – (10 – x )2
36 – x2 = 64 – 100 – x2 + 20 x
x = 3.6
From equation (i), AP2 = 36 – (3.6)2
AP = 4.8 cm
By circle’s property, PQ = 2 AP = 9.6 cm