**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} = OP^{2} + O’P^{2 }= 6^{2} + 8^{2 }= 100

OO’ = 10 cm

Let’s consider OA = x, therefore AO’ = 10 – x

Now in Δ POA, AP^{2} = OP^{2} – OA^{2} = 36 – x^{2} ……………….. (i)

Similarly, Now in Δ PO’A, AP^{2} = O’P^{2} – O’A^{2} = 64 – (10 – x)^{2} …………. (ii)

From equations (i) and (ii), we get:

36 – x^{2 }= 64 – (10 – x )^{2}

36 – x^{2 }= 64 – 100 – x^{2 }+ 20 x

x = 3.6

From equation (i), AP^{2} = 36 – (3.6)^{2}

AP = 4.8 cm

By circle’s property, PQ = 2 AP =** 9.6 cm**