Given equation of circles are 

S₁ = x² + y² + 2x + 6y = 0 ..................1

S₂ = x² + y² − 4x − 2y − 6 = 0 ............2

Subtract equation 1 - equation 2, we get

S₁ - S₂ = 6x + 8y + 6 = 0

=> 6x + 8y + 6 = 0 ............3

Cente of the circle S₂ is (2, 1)

Now, the length of the perpendicular from the center (2, 1) of of the circle 2 upon the common chord 3 is

      l = (6*2 + 8*1 + 6)/√{6² + 8² }

=> l = (12 + 8 + 6)/√{36 + 64}

=> l = 26/√100

=> l = 26/10

=> l = 13/5

Radius of the circle 2 is

      r = √{2² + 1² - (-6)}

=> r = √{4 + 1 + 6}

=> r = √11

Now length of common chord = 2√{r² - l² }

                                          = 2√{(√11)² - (13/5)² }

                                          = 2√{11 - 169/25}

                                          = 2√{(275 - 169)/25}

                                          = 2√{106/25} unit