circumcentre of equilateral triangle

Question

Given the circumcenter of an equilateral triangle how can i find the length of the side of the triangle?

The exact question is like this: An equilateral triangle with circumcenter at (-2, 5) having one side on the Y-axis, what will be the length of the side of the triangle?

I tried and tried but I could not get the traingle; instead I am getting a straight line. Can someone please explain the method to derive the answer to the above question?

Thanks

Answer 1

Ok I found the answer.

As pointed out in the comment above, the circumcenter of equilateral triangle will be same as the centroid. Also, with its $BC$ side on the $y$-axis, we have the three coordinates as under:

$A(x_1,y_1), B(0,y_2), C(0,y_3)$

With centroid at $G(-2, 5)$,

$x_G = (x_1+x_2+x_3)/3$

So $x_1$ will be $-6$.

Next I derived this from the distance formula ($AB^2 = AC^2$):

i.e. $6^2 + (y_2-y_1)^2 = 6^2 + (y_3-y_1)^2$

Solving this, we get $2y_1 = y_2 + y_3\tag{1}$

Now using formula of centroid for coordinate $y$,

$(y_1 + y_2 + y_3)/3 = 5\tag{2}$

Solving $(1)$ and $(2)$:

$y_1 = 5$

According to definition of circumcenter, the distance of circumcenter from any vertex of triangle would be same. So,

$GA^2 = GB^2$

Substituting the values and simplifying, we get

$y_2^2-10y_2+13=0$

On solving,

$y_2 = 5 \pm 2\sqrt3$

Substituting value of $y_1$ in $(1)$, and solving for $y_3$, we get

For $y_2 = 5 + 2\sqrt3$ $, \;\; y_3 = 5 - 2\sqrt3$

And for $y_2 = 5 - 2\sqrt3$ $, \;\; y_3 = 5 + 2\sqrt3$

And now using the distance formula for vertices $B$ and $C$, the length of the side of the triangle comes to $4\sqrt3$.

While this is the correct answer, I feel this is a bit too long and there must be a simpler and more elegant way of solving it.

Thanks