Advertisements
Advertisements
प्रश्न
Prove that the common chord of two intersecting circles is bisected at right angles by the line of centres.
Advertisements
उत्तर
Given: Two interesting circles with centres C &D.
AB is their common chord.
To prove: AB bisected by CD at right angles.
Proof : CA = CB ...(radii)
∴ C lies on the right bisector of AB.
Similarly, D lies on the right bisector of AB.
Therefore, CD is the right bisector of AB.
Hence proved.
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.
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 A and B.
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect each other at point P. Show that P is equidistant from the opposite sides AB and CD.
Draw a line AB = 6 cm. Draw the locus of all the points which are equidistant from A and B.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
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 the centres of all circles passing through two fixed points.
Sketch and describe the locus of the vertices of all triangles with a given base and a given altitude.
In Fig. ABCD is a quadrilateral in which AB = BC. E is the point of intersection of the right bisectors of AD and CD. Prove that BE bisects ∠ABC.
Given: ∠BAC, a line intersects the arms of ∠BAC in P and Q. How will you locate a point on line segment PQ, which is equidistant from AB and AC? Does such a point always exist?
