Question

Find the coordinates of circumcentre and radius of circumcircle of ∆ABC if A(7, 1), B(3, 5) and C(2, 0) are given.

Answer 1

Let the circumcentre be $\mathrm{P}\left(a,b\right)$.
The given points are A(7, 1), B(3, 5) and C(2, 0).
The circumcircle passes through the points A, B and C and the thus,
PA = PB = PC
$\Rightarrow {\mathrm{PA}}^2={\mathrm{PB}}^2={\mathrm{PC}}^2$
$$\begin{gathered} P{A}^2=P{B}^2 \\ \Rightarrow {\left(3-a\right)}^2+{\left(5-b\right)}^2={\left(7-a\right)}^2+{\left(1-b\right)}^2 \\ \Rightarrow 9+a^2-6a+25+b^2-10b=49+a^2-14a+1+b^2-2b \\ \Rightarrow a-b=2.....\left(1\right) \\ P{A}^2=P{C}^2 \\ \Rightarrow {\left(7-a\right)}^2+{\left(1-b\right)}^2={\left(2-a\right)}^2+{\left(0-b\right)}^2 \\ \Rightarrow 49+a^2-14a+1+b^2-2b=4+a^2-4a+b^2 \\ \Rightarrow 5a+b=23.....\left(2\right) \\ \left(1\right)+\left(2\right) \\ a=\frac{25}{6},b=\frac{13}{6} \end{gathered}$$
Radius = PC =
$$\begin{gathered} =\sqrt{{\left(\frac{25}{6}-2\right)}^2+{\left(\frac{13}{6}-0\right)}^2} \\ =\sqrt{{\left(\frac{13}{6}\right)}^2+{\left(\frac{13}{6}\right)}^2} \\ =\frac{13}{6}\sqrt{2} \end{gathered}$$