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:

Step 1: Since OP is O’P

OO’² = OP² + O’P²   = 6² + 8²  = 100

OO’ = 10 cm

Step 2: Now, we know that when two circles intersect each other, then the line joining their centers is the perpendicular bisector of the common cord.

∴ OO’ is perpendicular to  PQ and PA = AQ

Step 3: Let’s consider OA = x, therefore AO’ = 10 – x

Now in Δ POA, ∠ PAO = 90⁰

∴ OP² = PA² + OA²

∴ PA² = OP² – OA²

= 36 – x² ……………….. (i)

Step 4: Similarly, Now in Δ PO’A, ∠ PAO’ = 90⁰

∴ PA² = O’P² – O’A²

= 64 – (10 – x)² …………. (ii)

Step 5: From equations (i) and (ii), we get:

36 – x²  = 64 – (10 – x )²

36 – x²  = 64 – 100 – x²  + 20 x

x = 3.6

Step 6: From equation (i), PA² = 36 – (3.6)²

∴ PA = 4.8 cm

Step 7: Since PA = AQ = 4.8 cm

∴ PQ = PA + AQ = 4.8 + 4.8 = 9.6 cm

Therefore the length of PQ is 9.6 cm

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