Question

If two circles intersect in two points, prove that the line passing through their centre is the perpendicular bisectors of the common chord.

Answer 1

Let P and Q be the centres of two circles.
And let A and B be two points of intersection of the circles.
Let M be the point of intersection of AB and PQ.
In $△$APQ and $△$BPQ,
AP = BP ( radii of the same circle)
AQ = BQ ( radii of the same circle)
PQ = PQ (common side)

$$\begin{gathered} \therefore △\mathrm{APQ}\cong △\mathrm{BPQ}\left(\text{by}\mathrm{SSS}\mathrm{congruency}\mathrm{rule}\right) \\ \therefore \angle \mathrm{APM}=\angle \mathrm{BPM}-----\left(1\right) \\ \mathrm{Now},\mathrm{In}△\mathrm{APM}\mathrm{and}△\mathrm{BPM}, \\ \angle \mathrm{APM}=\angle \mathrm{BPM}\left[\mathrm{using}\left(1\right)\right] \\ \mathrm{PA}=\mathrm{PB}\left(\mathrm{radii}\mathrm{of}\mathrm{the}\mathrm{same}\mathrm{circle}\right) \\ \mathrm{PM}=\mathrm{PM}\left(\mathrm{common}\mathrm{side}\right) \\ \therefore △\mathrm{APM}\cong △\mathrm{BPM}\left(\text{by}\mathrm{SAS}\mathrm{congruency}\mathrm{rule}\right) \\ \mathrm{AM}=\mathrm{BM}(\mathrm{c}.\mathrm{p}.\mathrm{c}.\mathrm{t})--------\left(3\right) \\ \therefore \angle \mathrm{PMA}=\angle \mathrm{PMB}-----\left(2\right) \\ \mathrm{And}\angle \mathrm{PMA}+\angle \mathrm{PMB}=180°\left(\mathrm{angle}\mathrm{of}\mathrm{straight}\mathrm{line}\mathrm{is}180°\right) \\ \Rightarrow \angle \mathrm{PMA}+\angle \mathrm{PMA}=180°\left[\mathrm{using}\left(2\right)\right] \\ \Rightarrow \angle \mathrm{PMA}=90° \\ \therefore \angle \mathrm{PMA}=\angle \mathrm{PMB}=90°-----\left(4\right) \\ \mathrm{From}\left(3\right)\mathrm{an}\left(4\right),\mathrm{PQ}\hspace{0.17em}\mathrm{is}\text{the}\mathrm{perpendicular}\mathrm{bisector}\mathrm{of}\mathrm{AB}. \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$