From the given arc is known the start point the end point and a random point on the arc how we can find the center point coordinates? From the given arc is known the start point the end point and a random point
on the arc how we can find the center point coordinates?
 A: If the three points that you have are $A, B, C$, then construct the perpendicular bisectors of $AB$ and $BC$.  Where they intersect is the centre that you are looking for.
A: If you're after a geometric construction, then the first four sentences of André Nicolas' answer are exactly what you want.
If you want an algebraic construction, then write the three points as $(a_x,a_y), (b_x,b_y), (c_x,c_y)$ and the centre as $(x,y)$.  Then the distance formula gives you $(x-a_x)^2+(y-a_y)^2 = (x-b_x)^2+(y-b_y)^2=(x-c_x)^2+(y-c_y)^2$.  If you expand this out, the square terms all go away, leaving two simple simultaneous linear equations for $x$ and $y$.
The case where the simultaneous equations have no solution corresponds to the three given points lying on a straight line.
A: This point is called the circumcenter of the triangle formed by these three points.
Given $\triangle ABC$ with sides $a$, $b$, and $c$ opposite the corresponding vertices, the circumcenter of $\triangle ABC$ is
$$
\frac{a^2(b^2+c^2-a^2)A+b^2(c^2+a^2-b^2)B+c^2(a^2+b^2-c^2)C}{(a^2+b^2+c^2)^2-2(a^4+b^4+c^4)}\tag{1}
$$

Justification of $\mathbf{(1)}$:
The barycentric coordinates of $O$ are the proportions of the red, green, and blue triangles to the whole triangle. That is,
$$
O=\frac{\color{#C00000}{|\triangle BOC|}A+\color{#00A000}{|\triangle COA|}B+\color{#0000FF}{|\triangle AOB|}C}{|\triangle ABC|}\tag{2}
$$
$\hspace{4.5cm}$
Suppose we know the coordinates of $A$, $B$, and $C$, and that $|OA|=|OB|=|OC|=r$.
Since inscribed angles are half the corresponding central angles, $\left\{\small\begin{array}{ccc}\angle BAC=\angle BOC/2\\\angle CBA=\angle COA/2\\\angle ACB=\angle AOB/2\end{array}\right\}$ and
$$
\small\begin{array}{c|c|c}
a=2r\sin(\angle BAC)\quad&\quad\cos(\angle BAC)=\frac{b^2+c^2-a^2}{2bc}\quad&\quad|\triangle BOC|=\frac{r^2}{2}\sin(\angle BOC)\\
b=2r\sin(\angle CBA)\quad&\quad\cos(\angle CBA)=\frac{c^2+a^2-b^2}{2ca}\quad&\quad|\triangle COA|=\frac{r^2}{2}\sin(\angle COA)\\
c=2r\sin(\angle ACB)\quad&\quad\cos(\angle ACB)=\frac{a^2+b^2-c^2}{2ab}\quad&\quad|\triangle AOB|=\frac{r^2}{2}\sin(\angle AOB)\\
\end{array}
$$
Thus, we can compute quantities proportional to the areas of the red, green, and blue triangles
$$
\begin{align}
\frac{4abc}{r}|\triangle BOC|=2abcr\sin(\angle BOC)=a^2(b^2+c^2-a^2)\\
\frac{4abc}{r}|\triangle COA|=2abcr\sin(\angle COA)=b^2(c^2+a^2-b^2)\\
\frac{4abc}{r}|\triangle AOB|=2abcr\sin(\angle AOB)=c^2(a^2+b^2-c^2)
\end{align}\tag{3}
$$
Plugging $(3)$ into $(2)$ yields $(1)$.
