Question

Find the coordinates of the centre of the circle passing through the points P(6, –6), Q(3, –7) and R (3, 3).

Answer 1

Let the centre be P(a, b)
RO = PO = QO
${\mathrm{RO}}^2={\mathrm{PO}}^2$
$$\begin{gathered} \Rightarrow {\left(3-a\right)}^2+{\left(3-b\right)}^2={\left(6-a\right)}^2+{\left(-6-b\right)}^2 \\ \Rightarrow 9+a^2-6a+9+b^2-6b=36+a^2-12a+36+b^2+12b \\ \Rightarrow 6a-18b=54 \\ \Rightarrow a-3b=9.....\left(1\right) \end{gathered}$$
${\mathrm{PO}}^2={\mathrm{OQ}}^2$
$$\begin{gathered} {\left(6-a\right)}^2+{\left(-6-b\right)}^2={\left(3-a\right)}^2+{\left(-7-b\right)}^2 \\ \Rightarrow 36+a^2-12a+36+b^2+12b=9+a^2-6a+49+b^2+14b \\ \Rightarrow 3a+b=7.....\left(2\right) \end{gathered}$$
Multiplying (2) with 3 we get
$$\begin{gathered} a-3b=9.....\left(3\right) \\ 9a+3b=21.....\left(4\right) \\ \mathrm{Solving}\left(3\right)\mathrm{and}\left(4\right) \\ \mathrm{a}=3,\mathrm{b}=-2 \end{gathered}$$