The length of the common chord of the circle $x ^ { 2 } + y ^ { 2 } + 5 x + 7 y + 9 = 0$ and $x ^ { 2 } + y ^ { 2 } + 7 x + 5 y + 9 = 0$

The common chord of two circles S1=0 and S2=0 is given by S1 - S2 = 0. (x^2 + y^2 + 5x + 7y + 9) - (x^2 + y^2 + 7x + 5y + 9) = 0 => -2x + 2y = 0 => y = x. Substituting y=x into the first circle: x^2 + x^2 + 5x + 7x + 9 = 0 => 2x^2 + 12x + 9 = 0. This does not yield a standard chord length calculation without further steps. Re-calculating: The circles are x^2+y^2+5x+7y+9=0 and x^2+y^2+7x+5y+9=0. Subtracting gives 2x-2y=0, so y=x. The distance from the center (-2.5, -3.5) to the line x-y=0 is |(-2.5)-(-3.5)|/sqrt(2) = 1/sqrt(2). Radius^2 = 2.5^2 + 3.5^2 - 9 = 6.25 + 12.25 - 9 = 9.5. Chord length = 2 * sqrt(R^2 - d^2) = 2 * sqrt(9.5 - 0.5) = 2 * sqrt(9) = 6.