The length of the common chord of the circle $x^{2}+y^{2}+2x+3y+1=0$ and $x^{2}+y^{2}+4x+3y+2=0$ is
- $\frac{9}{2}$
- $2\sqrt{2}$
- $3\sqrt{2}$
- $\frac{3}{2}$
Subtract the two circle equations to find the common chord line: (x^2+y^2+4x+3y+2) - (x^2+y^2+2x+3y+1) = 0, which simplifies to 2x + 1 = 0 or x = -0.5. Substituting x = -0.5 into the first circle equation gives y^2 + 3y + 0.25 = 0, so y = (-3 +/- sqrt(9 - 1))/2 = (-3 +/- 2*sqrt(2))/2. The length of the chord is the difference between these y-values, which is 2*sqrt(2).
Subtracting the first circle equation from the second gives the equation of the common chord, 2x + 1 = 0, so the chord lies on the vertical line x = -1/2. The center of the second circle is (-2, -3/2) and its radius squared is 4 + 9/4 - 2 = 17/4. The perpendicular distance from the center to the line x = -1/2 is |-2 - (-1/2)| = 3/2. Using the Pythagorean theorem, the half-chord length is sqrt(17/4 - 9/4) = sqrt(8/4) = sqrt(2), making the total length of the chord 2 * sqrt(2) or 2 * sqrt(2).