1. If a number of circles touch a given line segment PQ at a point A, then their

centres lie on the perpendicular bisector of PQ.

2. 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.

3. AB is a diameter of a circle and AC is its chord such that ∠BAC = 30°. If the

tangent at C intersects AB extended at D, then BC = BD.

1> Given: PQ is the line segment.

Let a circle C (O, r) touches PQ at A

.

∴ ∠OAQ = 90° (Radius is perpendicular to the tangent at point of contact)

Let C (O', r') be another circle with touches PQ at A.

∴ ∠O'AQ = 90° (Radius is perpendicular to the tangent at point of contact)

This is possible, when O and O' lie on the same line OO'A.

⇒the centres of all circle touching PQ at A lie on the perpendicular to PQ at A.

2> we have the following situation-

Let A and B be the centres of the two circles passing through the end points of line segment PQ.

In triangle ACP and ACQ

AP = AQ

AC = AC

Hence CP = CQ (corresponding part of congruent triangle)

So centres lie on the perpendicular bisector of PQ.

3> We have-

 (angle made by the segment on centre is double to that of the any part of the circle)

In triangle OCD,

(angle sum property of triangle)

Now, 

(exterior angle property)

Hence