# Question #608f0

Mar 17, 2015

The answer is: $D \left(\frac{3}{2} , \frac{7}{2}\right)$.

The point we are searching is the circumcentre, that is the centre of the circumcircle. It is the point at which the perpendicular bisectors of the sides of the triangle are concurrent.

We can find it finding the axis of two sides. The axis of a segment is the perpendicular line that passes from the middle of the segment. It is the geometrical locus of the equidistant points from the vertex of the segment.

This is the graph:

Let $P \left(x , y\right)$ be a generic point of the plane.

$A P = C P \Leftrightarrow$

$\sqrt{{\left({x}_{p} - {x}_{a}\right)}^{2} + {\left({y}_{p} - {y}_{a}\right)}^{2}} = \sqrt{{\left({x}_{p} - {x}_{c}\right)}^{2} + {\left({y}_{p} - {y}_{c}\right)}^{2}}$

$\sqrt{{\left(x + 3\right)}^{2} + {\left(y + 3\right)}^{2}} = \sqrt{{\left(x - 9\right)}^{2} + {\left(y - 1\right)}^{2}}$

${x}^{2} + 6 x + 9 + {y}^{2} + 6 y + 9 = {x}^{2} - 18 x + 81 + {y}^{2} - 2 y + 1$

$8 y = - 24 x + 64 \Rightarrow y = - 3 x + 8$

$C P = B P \Leftrightarrow$

$\sqrt{{\left({x}_{p} - {x}_{c}\right)}^{2} + {\left({y}_{p} - {y}_{c}\right)}^{2}} = \sqrt{{\left({x}_{p} - {x}_{b}\right)}^{2} + {\left({y}_{p} - {y}_{b}\right)}^{2}}$

$\sqrt{{\left(x - 9\right)}^{2} + {\left(y - 1\right)}^{2}} = \sqrt{{\left(x + 5\right)}^{2} + {\left(y - 8\right)}^{2}}$

${x}^{2} - 18 x + 81 + {y}^{2} - 2 y + 1 = {x}^{2} + 10 x + 25 + {y}^{2} - 16 y + 64$

$14 y = 28 x + 7 \Rightarrow y = 2 x + \frac{1}{2}$

And now, let's solve the system:

$y = - 3 x + 8$
$y = 2 x + \frac{1}{2}$

$- 3 x + 8 = 2 x + \frac{1}{2} \Rightarrow 5 x = 8 - \frac{1}{2} \Rightarrow x = \frac{1}{5} \cdot \frac{16 - 1}{2} \Rightarrow$

$x = \frac{3}{2}$
$y = 2 \cdot \frac{3}{2} + \frac{1}{2} = \frac{7}{2}$

So the point is: $D \left(\frac{3}{2} , \frac{7}{2}\right)$.