The length of common chord of circles $x^{2}+y^{2}+2x+6y=0$ & $x^{2}+y^{2}-4x-2y-6=0$ is $\dfrac{\alpha\sqrt{106}}{\beta}$ where $\alpha$ and $\beta$ are coprime then

Subtracting the two circle equations gives the common chord, 6x + 8y + 6 = 0, or 3x + 4y + 3 = 0. The center of the first circle is (-1, -3) and its radius squared is 10. The perpendicular distance from the center to the chord is d = |-3 - 12 + 3| / 5 = 12/5. The half-chord length is sqrt(10 - 144/25) = sqrt(106/25) = sqrt(106)/5, making the full chord length 2 * sqrt(106)/5. Here alpha = 2 and beta = 5, so alpha - beta = -3 and the absolute difference is 3.

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