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.
Construct a right angled triangle PQR, in which ∠Q = 90°, hypotenuse PR = 8 cm and QR = 4.5 cm. Draw bisector of angle PQR and let it meets PR at point T. Prove that T is equidistant from PQ and QR.
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.
Draw an angle ABC = 75°. Draw the locus of all the points equidistant from AB and BC.
In the given figure, obtain all the points equidistant from lines m and n; and 2.5 cm from O.
In a quadrilateral PQRS, if the bisectors of ∠ SPQ and ∠ PQR meet at O, prove that O is equidistant from PS and QR.
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.
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.
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?
The bisectors of ∠B and ∠C of a quadrilateral ABCD intersect in P. Show that P is equidistant from the opposite sides AB and CD.
