Question

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.

Answer 1

Draw the largest among all the circles of any radii having center O, and draw a line segment P Q, touching the circle at point A.

Now, draw, as many circles as you can having P Q, as a tangent, and the tangent P Q, touches all the circles at Point A.

Consider the first circle , having center O.

Join OP and O Q.

As line joining from center to point of contact of tangent ,is perpendicular to tangent.

∠OAP=∠OAQ=90°

Now, we have to prove that , PO=OQ

In Δ OAP and ΔOAQ

∠OAP=∠OAQ→→Each being 90°

OA is common.

PA=AQ→→[Given]

Δ OAP ≅ ΔOAQ→→[SAS]

OP=OQ→→[CPCT]

Similarly, A'P=A'Q

BP=B Q

Shows that center of all circles  lies on perpendicular bisector.

Now,  If OA is perpendicular bisector of PQ

Then, PA=AQ and OP=OQ,→→Perpendicular bisector divides the line segment in two equal parts, as well as make an angle of 90°, at the point where it touches the line segment.

As,  line joining from center to point of contact of tangent ,is perpendicular to the tangent.

So, the center of all the circles will lie on line AO.