Question

State true or false

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

• A. True
• B. False

Answer 1

The correct option is A True

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

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

Consider the first circle , having center $O$.

Join $OP$ and $OQ$.

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

$\angle OAP=\angle OAQ=90°$

Now, we have to prove that , $PO=OQ$

In $\Delta OAP$ and $\Delta OAQ$

$\angle OAP=\angle OAQ$→→Each being $90°$

$OA$ is common.

$PA=AQ$→→[Given $A$ is mid-point]

$\Delta OAP\cong \Delta OAQ$→→[SAS]

$OP=OQ$→→[CPCT]

Similarly, $A^{\prime }P=A^{\prime }Q$

$BP=BQ$

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 $AO$.