If a number of circles touch a given line segment PQ at a point A, then their centers lie on the perpendicular bisector of PQ.

Question

Philoid · Accepted Answer

False

Let the S1, S2, S3, …., Sn be n circles with centers C1, C2, C3, …, Cn respectively.

And The PQ is a common tangent to all the circles at point A which is common to all circles.

As we know,

tangent at any point on the circle is perpendicular to the radius through point of contact

we have,

C1A ⏊ PQ

C2A ⏊ PQ

C3A ⏊ PQ

CnA ⏊ PQ

So, C1 C2 C3 … Cn all lie on the perpendicular line to PQ but not on perpendicular bisector as

PA may or may not be equal to AQ .