If a number of circles pass through the end points P and Q of a line segment PQ, then their centres lie on the perpendicular bisector of PQ.

Explanation

Given that PQ is any line segment and `S_(1),S_(2),S_(3),S_(4),.....` circles are touch a line segment PQ at a point A. Let the centres of the circles `S_(1),S_(2),S_(3),S_(4),.....` be `C_(1),C_(2),C_(3),C_(4).....` respectively.

To prove Centres of these circles lie on the perpendicular bisector of PQ. Now, jonining each centre of the circles to the point A on the line segment PQ by a line segment i.e., `C_(1)A,C_(2)A,C_(3)A,C_(4)A....` so on.
We know that, if we draw a line from the centre of a circle to its tangent line, then the line is always perpendicular to the tangent line. But it not bisect the line segment PQ
So,`C_(1)AbotPQ` `[forS_(1)]`
`C_(2)AbotPQ` `[forS_(2)]`
`C_(3)AbotPQ` `[forS_(3)]`
`C_(4)AbotPQ` `[forS_(4)]`
.....so on.
Since, each circle is passing through a point A. Therefore, all the line segments `C_(1)A,C_(2)A,C_(3)A,C_(4)A,.....` so on are coincident.
So centre of each circle lies on the perpendicular line of PQ but they do not lie on the perpendicular bisector of PQ.
Hence, a number fo circles touch a given line segment PQ at a point A, then their centres lie