Multiple choice

The length of the common chord of the circles: $x^{2}+y^{2}+6x+5=0$ and $x^{2}+y^{2}+4y-5=0$ is

  1. $\sqrt{\dfrac{12}{13}}$
  2. $\displaystyle \frac{12}{\sqrt{13}}$
  3. $\displaystyle \frac{\sqrt{12}}{13}$
  4. $\sqrt{\dfrac{13}{12}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Common chord equation: (x^2+y^2+6x+5) - (x^2+y^2+4y-5) = 0 => 6x - 4y + 10 = 0 => 3x - 2y + 5 = 0. Distance from center of first circle (-3, 0) to line: |3(-3) - 2(0) + 5| / sqrt(3^2 + (-2)^2) = |-4| / sqrt(13) = 4/sqrt(13). Radius of first circle = sqrt(3^2 + 0^2 - 5) = 2. Half-length of chord = sqrt(r^2 - d^2) = sqrt(4 - 16/13) = sqrt(36/13) = 6/sqrt(13). Total length = 12/sqrt(13).

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