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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
Solution
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.
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