Answer to Question #139729 in Analytic Geometry for Aravind

Question

The coordinates of the vertices of a triangle are A( x1, y1), B( x2, y2), and C(x3,y3) .Then what's the co ordinates of the circumcentre.

Answer 1

$A(x_1,y_1), \ B(x_2,y_2),\ C(x_3,y_3)$

To calculate the coordinate of the circumcenter, we have to solve any 2 bisector equations of any of the three lines AB, BC, or AC, then find out the intersection points. i.e.

The midpoint of side AB = $\lgroup \dfrac{x_1 + x_2}{2}, \dfrac{y_1 + y_2}{2} \rgroup= (x_{12}, y_{12})$

And the slope of AB = $\dfrac{y_2-y_1}{x_2-x_1} = m_{12}$

$\therefore$ The slope of the perpendicular bisector of side AB = $\frac{-1}{m_{12}} = m_{21}$

The equation of the perpendicular bisector of AB is therefore,

$y-y_{12} = m_{21}(x-x_{12}) \ ---(i)$

Similarly,

For side AC

The midpoint of side AC =

$\lgroup \dfrac{x_1 + x_3}{2}, \dfrac{y_1 + y_3}{2} \rgroup= (x_{13}, y_{13})$

And its slope = $\dfrac{y_3-y_1}{x_3-x_1} = m_{13}$

Making the slope of its perpendicular bisector = $\frac{-1}{m_{13}} = m_{31}$

The equation of the perpendicular bisector of length AC is therefore,

$y-y_{13} = m_{31}(x-x_{13}) \ ---(ii)$

On solving equations (i) and (ii), the values of x and y is the coordinate of the circumcenter

OR, using formula method;

$Circumcenter(X, Y) = \lgroup \dfrac{x_1 sin 2A + x_2 sin 2B + x_3 sin 2C}{sin 2A + sin 2B + sin 2C},\dfrac{y_1 sin 2A + y_2 sin 2B + y_3 sin 2C}{sin 2A + sin 2B + sin 2C}\rgroup$