Find the coordinates of circumcenter of a tri.ABC where A(1,2) , B(3,-4) , C(5,-6).
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Hi, 
Let the coordinates of the circumcentre of the triangle be (x, y).
Circumcentre of a triangle is equidistant from each of the vertices.

Distance between (1,2) and (x, y) = Distance between (3,-4) and (x, y)
$$\begin{gathered} \sqrt{{\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-2\right)}^2}=\sqrt{{\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}+4\right)}^2} \\ \mathrm{square} \\ {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-2\right)}^2={\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}+4\right)}^2 \\ {\mathrm{x}}^2-2\mathrm{x}+1+{\mathrm{y}}^2+4-4\mathrm{y}= {\mathrm{x}}^2+9-6\mathrm{x}+{\mathrm{y}}^2+16+8\mathrm{y} \\ 4\mathrm{x}-12\mathrm{y}-20 = 0 \\ \mathrm{x}-3\mathrm{y} = 5 ....\left(1\right) \\ \mathrm{also} \mathrm{distance} \mathrm{between} \left(3,-4\right) \mathrm{and} \left(\mathrm{x},\mathrm{y}\right)= \mathrm{distance} \mathrm{between} \left(5,-6\right) \mathrm{and} \left(\mathrm{x},\mathrm{y}\right) \\ \sqrt{{\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}+4\right)}^2}=\sqrt{{\left(\mathrm{x}-5\right)}^2+{\left(\mathrm{y}+6\right)}^2} \\ \mathrm{square} \\ {\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}+4\right)}^2={\left(\mathrm{x}-5\right)}^2+{\left(\mathrm{y}+6\right)}^2 \\ {\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2+16+8\mathrm{y}= {\mathrm{x}}^2+25-10\mathrm{x}+{\mathrm{y}}^2+36+12\mathrm{y} \\ 4\mathrm{x}-4\mathrm{y}=36 \\ \mathrm{x}-\mathrm{y} = 9 ....\left(2\right) \\ \mathrm{subtract} \mathrm{equation} 1 \mathrm{and} 2 \\ 2\mathrm{y} = 4 \\ \mathrm{y} = 2 \mathrm{so} \mathrm{putting} \mathrm{this} \mathrm{in} \mathrm{any} \mathrm{equation} \\ \mathrm{x} = 9+2 = 11 \end{gathered}$$