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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
The given figure shows a triangle ABC in which AD bisects angle BAC. EG is perpendicular bisector of side AB which intersects AD at point F.
Prove that:
F is equidistant from AB and AC.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
Describe the locus for questions 1 to 13 given below:
1. The locus of a point at a distant 3 cm from a fixed point.
Describe the locus of points at a distance 2 cm from a fixed line.
Describe the locus of the moving end of the minute hand of a clock.
Describe the locus of a stone dropped from the top of a tower.
Describe the locus of points at distances less than 3 cm from a given point.
By actual drawing obtain the points equidistant from lines m and n; and 6 cm from a point P, where P is 2 cm above m, m is parallel to n and m is 6 cm above n.
In a quadrilateral ABCD, if the perpendicular bisectors of AB and AD meet at P, then prove that BP = DP.
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
