Question

Find the coordinates of the circumcentre of a triangle whose vertices are (–3,1), (0,–2) and (1,3)

Answer 1

Let the given vertices A(–3, 1), B(0, –2) and C(1, 3).
Circumcircle passes through the vertices of a triangle.
Let the circumcentre be P(a, b)
The distance of point P will be equal from A, B and C.
PA = PB = PC
$\Rightarrow$${\mathrm{PA}}^2={\mathrm{PB}}^2={\mathrm{PC}}^2$
$$\begin{gathered} {\mathrm{PB}}^2={\mathrm{PC}}^2 \\ \Rightarrow {\left(1-a\right)}^2+{\left(3-b\right)}^2={\left(0-a\right)}^2+{\left(-2-b\right)}^2 \\ \Rightarrow 1+a^2-2a+9+b^2-6b=a^2+4+b^2-2\times b\times \left(-2\right) \\ \Rightarrow 10+a^2-2a+9+b^2-6b=a^2+4+b^2+4b \\ \Rightarrow a+5b=3.....\left(1\right) \\ {\mathrm{PB}}^2={\mathrm{PA}}^2 \\ \Rightarrow {\left(a-0\right)}^2+{\left(b+2\right)}^2={\left(a+3\right)}^2+{\left(b-1\right)}^2 \\ \Rightarrow a^2+9+6a+b^2+1-2b=a^2+b^2+4+4b \\ \Rightarrow a-b=1.....\left(2\right) \end{gathered}$$
solving (1) and (2) we get
$a=\frac{-1}{3},b=\frac{2}{3}$
Thus, the circumcentre of the triangle is $\mathrm{P}\left(\frac{-1}{3},\frac{2}{3}\right)$