Question

Find the co-ordinates of the centre of a circle passing through the points A(3,5) B($-$1, 2) and C(3, $-$3). Also find the radius of circle.

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{coordinate}\text{s}\mathrm{of}\mathrm{circumcentre}\mathrm{O}\mathrm{be}\left(\mathrm{x},\mathrm{y}\right). \\ \mathrm{Distance}\mathrm{of}\mathrm{circumcentre}\mathrm{from}\mathrm{the}\mathrm{vertices}\mathrm{A}\left(3,5\right),\mathrm{B}\left(-1,2\right)\mathrm{and}\mathrm{R}\left(3,-3\right)\mathrm{of}\mathrm{the}\mathrm{triangle} \\ \mathrm{will}\mathrm{be}\mathrm{equal}\mathrm{as}\mathrm{it}\mathrm{is}\mathrm{equal}\mathrm{to}\mathrm{the}\mathrm{radius}\mathrm{of}\mathrm{the}\mathrm{circle}. \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{OA}=\mathrm{OB} \\ \Rightarrow {\mathrm{OA}}^2={\mathrm{OB}}^2 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-5\right)}^2={\left[x-\left(-1\right)\right]}^2+{\left(y-2\right)}^2 \\ \Rightarrow {\left(x-3\right)}^2+{\left(y-5\right)}^2={\left(x+1\right)}^2+{\left(y-2\right)}^2 \\ \Rightarrow x^2+9-6x+y^2+25-10y=x^2+1+2x+y^2+4-4y \\ \Rightarrow -6x+34-10y=2x+5-4y \\ \Rightarrow -8x-6y=-29 \\ \Rightarrow 8x+6y=29----\left(1\right) \end{gathered}$$

$$\begin{gathered} \therefore \mathrm{OB}=\mathrm{OC} \\ \Rightarrow {\mathrm{OB}}^2={\mathrm{OC}}^2 \\ \Rightarrow {\left[x-\left(-1\right)\right]}^2+{\left(y-2\right)}^2={\left(x-3\right)}^2+{\left[y-\left(-3\right)\right]}^2 \\ \Rightarrow {\left(x+1\right)}^2+{\left(y-2\right)}^2={\left(x-3\right)}^2+{\left(y+3\right)}^2 \\ \Rightarrow x^2+1+2x+y^2+4-4y=x^2+9-6x+y^2+9+6y \\ \Rightarrow 2x-4y+5=-6x+6y+18 \\ \Rightarrow 8x-10y=13----\left(2\right) \end{gathered}$$

$$\begin{gathered} \mathrm{Subtracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}: \\ 8x+6y=29 \\ 8x-10y=13 \\ \underline{-+-} \\ 16y=16 \\ \Rightarrow y=1 \\ \mathrm{Substituting}\mathrm{the}\mathrm{value}\mathrm{of}y\mathrm{in}\mathrm{equation}\left(1\right),\mathrm{we}\mathrm{get}: \\ 8x+6\times 1=29 \\ \Rightarrow 8x=23 \\ \Rightarrow y=\frac{23}{8} \\ \mathrm{Therefore},\left(\frac{23}{8},1\right)\text{are}\mathrm{the}\mathrm{coordinate}s\mathrm{of}\mathrm{the}\mathrm{circumcentre}\mathrm{of}\mathrm{triangle}\mathrm{PQR} \end{gathered}$$

$$\begin{gathered} \mathrm{Now},\mathrm{radius}=\mathrm{OA} \\ =\sqrt{{\left(\frac{23}{8}-3\right)}^2+{\left(1-5\right)}^2} \\ =\sqrt{{\left(\frac{-1}{8}\right)}^2+{\left(-4\right)}^2} \\ =\sqrt{\frac{1}{64}+16} \\ =\sqrt{\frac{1025}{64}} \\ =\frac{5\sqrt{41}}{8}\mathrm{units} \end{gathered}$$