Consider the fact that if P(x,y) is the unknown point, then when you apply rotation around that point, the point P1(x1, y1) goes to (gets mapped to) P2(x2,y2). From this you can construct equations and solve them to get the x and y. Just lookup the formulas for rotation in the 2D plane. The rotation angles are nice in both cases: 90 and 60 degrees. If you do this, you should get the answers you have posted here.
Say you rotate the point (x1,y1) around the point (a,b) on an angle of $\theta$.
Then the point (x1,y1) gets mapped to (x2, y2) where:
$x_2 = (x_1-a) * cos(\theta) - (y_1-b) * sin(\theta) + a$
$y_2 = (x_1-a) * sin(\theta) + (y_1-b) * cos(\theta) + b$
In your case you don't know the $a$ and $b$, you want to find them.
The $\theta$ is $\pm \pi/2$ and $\pm \pi/3$ respectively.
The $x_1, y_1, x_2, y_2$ - these you know.
Here is how one can solve the case: $\pi/3$. You have 3 more cases to solve.
See also:
Rotation
Rotation matrix
Consider this image:
The distance from the center of the triangle (a.k.a. apothem) to the midpoint of one of its side is $h/3$, where $h$ is the height of the triangle. So the radius of the circle r is the distance from the center of the triangle to a vertex, and this distance is $2h/3$.
Finding the side of the triangle (L)
We know that the area of the triangle is $7\sqrt{3}$, so we do:
$Area = \frac{1}{2}(base)(height) = \frac{1}{2}L\left(\frac{L\sqrt{3}}{2}\right) = 7\sqrt{3}$
Solving for L, we get $L = 2\sqrt{7}$.
Finding the radius of the circle (r)
As you already calculated, we can use the pythagorean theorem to find the height of the triangle:
$L^2 = \left(\frac{L}{2}\right)^2 + h^2$
$h = \sqrt{L^2 - \frac{L^2}{4}} = \frac{L\sqrt{3}}{2}$
The radius of the circle is then:
$r = \frac{2}{3}h = \frac{2}{3}\frac{L\sqrt{3}}{2} = \frac{L\sqrt{3}}{3} = \frac{2\sqrt{7}\sqrt{3}}{3} = \frac{2\sqrt{21}}{3}$
Finding the area of the circle
$Area_{circle} = \pi r^2 = \pi \left(\frac{2\sqrt{21}}{3}\right)^2 = \pi \frac{(4)(21)}{9} = \frac{28\pi}{3} \approx \frac{88}{3}$
Best Answer
First, find the length of a side of the triangle. $$l=\sqrt{(10-0)^2 +(-4-6)^2}=10\sqrt{2}$$ Next, we draw two circles of radius $l$ and centers at the given points. The intersection of these circles will give us the third point (well, there will be two points so two solutions to this problem). We have two equations: $$(x-10)^2+(y+4)^2=200$$ and $$x^2+(y-6)^2=(x-10)^2+(y+4)^2$$ From the second equation we have $x^2+y^2-12y+36=x^2-20x+100+y^2+8y+16$ which will give us $x=y+4$. Now substitute $x$ in the first equation to get a quadratic equation for $y$. $$(y-6)^2+(y+4)^2=2y^2+4y+52=200$$ This has two solutions, $y_1=-1-5\sqrt{3}$ and $y_2=-1+5\sqrt{3}$. Thus, we have two possible locations $(3-5\sqrt{3}, -1-5\sqrt{3})$ and $(3+5\sqrt{3}, -1+5\sqrt{3})$ for the third vertex.