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
There are many ways to find your desired point. Here is one way. Since you calculated a length as an approximation rather than exactly, I'll also use approximations for irrational values.
First, let's find the distance $d$ of the desired point from our "base point" $(2,3)$. The base point $(2,3)$, the midpoint of the line segment between the two given points, and the desired point form a $30°$-$60°$-$90°$ triangle. The short leg has half the length of the line segment that you already calculated. Let's call that $\frac L2$, where $L$ is your calculated length. We then use the $30°$ angle in the right triangle to find the desired distance $d$:
$$\cos 30°=\frac{\frac L2}{d}$$
$$d=\frac L{2\cos 30°}=1.82574185835$$
Now we want to find the angle of inclination from the base point to the desired point. The angle of inclination from $(2,3)$ to $(5,2)$ is $\tan^{-1}\frac{-1}3=-18.4349488229°$. We add $30°$ to that to get our desired angle of inclination $\theta=11.5650511771°$.
We now find the desired point by its position relative to the base point.
$$\begin{align}
(x,y) &= (2+d\cos\theta,3+d\sin\theta) \\
&= (3.78867513459,3.36602540378)
\end{align}$$
This answer checks in Geogebra.
As I state in a comment to this answer, the method above for finding $d$, the distance of the desired point from the base point, works only for isosceles triangles where you know the endpoints of the base as well as the angles there. If the angles at the two known vertices differ, there is another way.
Let's say the length of the known triangle side is $L$, the known angle at the base point is $\alpha$ and the angle at the other known vertex is $\beta$. We can use the law of sines to get
$$\frac{\sin(180°-\alpha-\beta)}{L}=\frac{\sin(\beta)}{d}$$
so
$$d=\frac{L\sin(\beta)}{\sin(180°-\alpha-\beta)}$$
The rest of my method above works after this. This more general method also works and checks in your particular problem. Let me know if you want me to edit my answer above to only show the more general method: that might be easier for you.
Best Answer
Call the position of point $C$ by the coords $(a, b)$. The equations for $C$ are then
$$ \sqrt{(a-3)^2 + (b - 4)^2} = \sqrt{26} \\ \sqrt{(a+2)^2 + (b - 3)^2} = \sqrt{26} $$ Squaring both, we get $$ (a-3)^2 + (b - 4)^2 = 26 \\ (a+2)^2 + (b - 3)^2 = 26 $$ $$ a^2 - 6a + 9 + b^2 - 8b + 16= 26 \\ a^2 + 4a + 4 + b^2 - 6b + 9= 26 $$ Subtracting these two gives $$ -10a + 5 - 2b + 7 = 0 $$ or $$ 6 = 5a + b $$ which is a line both points must lie on. Writing this as $$ b = 6 - 5a $$ we can substitute in either equation. Let's got with the second:
$$ a^2 + 4a + 4 + b^2 - 6b + 9= 26 $$ becomes $$ a^2 + 4a + 4 + (6-5a)^2 - 6(6-5a) + 9= 26 $$ which is a quadratic that can now be solved for the two possible values of $a$.
(Once you do so, you use $b = 6 - 5a$ to find the corresponding $b$-values.)