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.

Options

• True

• False

Answer 1

This statement is False.

Explanation:

Given that PQ is any line segment and S₁, S₂, S₃, S₄,... circles are touch a line segment PQ at a point A.

Let the centres of the circles S₁, S₂, S₃, S₄,... be C₁, C₂, C₃, C₄,... respectively.

To prove centres of these circles lie on the perpendicular bisector PQ.

Now, joining each centre of the circles to the point A on the line segment PQ by a line segment i.e., C₁A, C₂A, C₃A, C₄A... 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,

C₁A ⊥ PQ   ...[For S₁]

C₂A ⊥ PQ   ...[For S₂]

C₃A ⊥ PQ   ...[For S₃]

C₄A ⊥ PQ   ...[For S₄]

Since, each circle is passing through a point A.

Therefore, all the line segments C₁A, C₂A, C₃A, C₄A.... 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 of circles touch a given line segment PQ at a point A, then their centres lie on the perpendicular of PQ but not on the perpendicular bisector of PQ.