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प्रश्न
Show that the locus of the centres of all circles passing through two given points A and B, is the perpendicular bisector of the line segment AB.
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उत्तर
Let P and Q be the centres of two circles S and S', each passing through two given points A and B. Then,
PA = PB ...[Radii of the same circle]
⇒ P lies on the perpendicular bisector of AB ...(i)
Again, QA = QB ...[Radii of the same circle]
⇒ Q lies on the perpendicular bisector of AB ...(ii)
From (i) and (ii), it follows that P and Q both lies on the perpendicular bisector of AB.
Hence, the locus of the centres of all the circles passing through A and B is the perpendicular bisector of AB.
संबंधित प्रश्न
In each of the given figures; PA = PB and QA = QB.
| i. |  |
| ii. |  |
Prove, in each case, that PQ (produce, if required) is perpendicular bisector of AB. Hence, state the locus of the points equidistant from two given fixed points.
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- Point A is equidistant from all the three sides of the triangle.
- AM bisects angle LMN.
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- P is equidistant from B and C.
- P is equidistant from AB and BC.
- Measure and record the length of PB.
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- B and D.
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