Advertisements
Advertisements
प्रश्न
Describe the locus of points inside a circle and equidistant from two fixed points on the circumference of the circle.
Advertisements
उत्तर
The locus of the points inside the circle which are equidistant from the fixed points on the circumference of a circle will be the diameter which is perpendicular bisector of the line joining the two fixed points on the circle.
APPEARS IN
संबंधित प्रश्न
Construct a triangle ABC, in which AB = 4.2 cm, BC = 6.3 cm and AC = 5 cm. Draw perpendicular bisector of BC which meets AC at point D. Prove that D is equidistant from B and C.
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.
In the given triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Construct a triangle ABC, with AB = 7 cm, BC = 8 cm and ∠ABC = 60°. Locate by construction the point P such that:
- P is equidistant from B and C.
- P is equidistant from AB and BC.
- Measure and record the length of PB.
Describe the locus of points at distances less than 3 cm from a given point.
Describe the locus of points at distances greater than 4 cm from a given point.
In Δ ABC, the perpendicular bisector of AB and AC meet at 0. Prove that O is equidistant from the three vertices. Also, prove that if M is the mid-point of BC then OM meets BC at right angles.
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
ΔPBC and ΔQBC are two isosceles triangles on the same base. Show that the line PQ is bisector of BC and is perpendicular to BC.
ΔPBC, ΔQBC and ΔRBC are three isosceles triangles on the same base BC. Show that P, Q and R are collinear.
