Advertisements
Advertisements
प्रश्न
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
Advertisements
उत्तर
Let a circle of radius r (with centre B) touch a circle of radius R at C. Then ACB is a straight line and
AB = AC + CB = R + r
Thus, B moves such that its distance from fixed point. A remains constant and is equal to R + r.
Hence, the locus of B is a circle whose centre is A and radius equal to R + r.
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 triangle ABC, find a point P equidistant from AB and AC; and also equidistant from B and C.
Describe the locus of the centre of a wheel of a bicycle going straight along a level road.
Describe the locus of points at distances greater than or equal to 35 mm 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.
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.


