Question

(TRUE OR FALSE)
If a number of circles pass through the end points P and Q of a line segment PQ, then their centers lie on the perpendicular bisector of PQ .

Answer 1

Answer:

True!

Let as draw a line segment PQ and S1 S2 S3 …. Sn are n circles with center C1 C2 C3 … Cn respectively passing through p and Q

Here PQ either will be a chord to a circle or the diameter (possible only once when diameter length is equal to the length of PQ.

Now, if PQ is diameter then clearly center lies on the half of diameter i.e. on the perpendicular bisector of PQ.

But if PQ is a chord then we know that perpendicular from the center to the chord of circle bisect the chord implies that center lies on the perpendicular bisector of the chord.

So for each circle center lies on the perpendicular bisector of PQ.

So above statement is true .

Answer 2

TRUE

Step-by-step explanation:

Given that PQ is any line segment and $S_1,S_2, S_3, S_4,$,… circles are touch a line segment PQ at a point A.

Let the centres of the circles be $S_1,S_2, S_3, S_4,$ ...$C_1, C_2, C_3, C_4$ 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., $C_1A, C_2A, C_3A, C_4A$… 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.

$C_1 A \perp PQ\\\\C_2 A \perp PQ\\\\C_3 A \perp PQ\\\\C_4 A \perp PQ....$

Since, each circle is passing through a point A. Therefore, all the line segments  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.

#Learn more :

Draw a circle of radius 2 centimetre draw a diameter pq of the circle now construct of perpendicular bisector of line segment pq does it pass through the centre of circle

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