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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?

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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.

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With this compass and ruler method and the coordinate geometry approach, I think most bases are covered. (+1) – robjohn Jul 27 '12 at 20:29

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.

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Unfortunately, André deleted his answer. I'm not sure why. – robjohn Jul 27 '12 at 14:23
OK, I'll post the geometric answer too. – user22805 Jul 27 '12 at 19:33

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, $$ O=\frac{\color{#C00000}{|\triangle BOC|}A+\color{#00A000}{|\triangle COA|}B+\color{#0000FF}{|\triangle AOB|}C}{|\triangle ABC|}\tag{2} $$

$\hspace{4.5cm}$enter image description here

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 $(1)$.

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