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