Multiple choice

Find the circumcentre of the triangle whose vertices are $(0, 0)$, $\left ( 3, \sqrt{3} \right )$ and $\left ( 0, 2\sqrt{3} \right )$.

  1. $\left ( 1, \sqrt{3} \right )$
  2. $\left ( -1, \sqrt{3} \right )$
  3. $\left ( 0, \sqrt{3} \right )$
  4. $\left ( 1, 2\sqrt{3} \right )$
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A Correct answer
Explanation

The circumcenter is equidistant from (0,0), (3, sqrt(3)), and (0, 2*sqrt(3)). Let the center be (x,y). x^2 + y^2 = (x-3)^2 + (y-sqrt(3))^2 and x^2 + y^2 = x^2 + (y-2*sqrt(3))^2. From the second, y^2 = y^2 - 4*sqrt(3)*y + 12, so 4*sqrt(3)*y = 12, y = sqrt(3). Substituting into the first, x^2 + 3 = (x-3)^2 + 0, x^2 + 3 = x^2 - 6x + 9, 6x = 6, x = 1.

AI explanation

Let the circumcentre be (x,y), which is equidistant from all three vertices (0,0), (3, sqrt(3)) and (0, 2*sqrt(3)). Equating the squares of the distances from (0,0) and (3, sqrt(3)) gives x^2 + y^2 = (x-3)^2 + (y-sqrt(3))^2, which simplifies to 3x + sqrt(3)y = 6. Equating distances from (0,0) and (0, 2*sqrt(3)) yields x^2 + y^2 = x^2 + (y - 2*sqrt(3))^2, simplifying to y = sqrt(3). Substituting this back into the first equation gives 3x = 3, so x = 1, making the circumcentre (1, sqrt(3)).

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