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प्रश्न
Find the locus of the centre of a circle of radius r touching externally a circle of radius R.
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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 triangle LMN, bisectors of interior angles at L and N intersect each other at point A. Prove that:
- Point A is equidistant from all the three sides of the triangle.
- AM bisects angle LMN.
Use ruler and compasses only for this question.
- Construct ΔABC, where AB = 3.5 cm, BC = 6 cm and ∠ABC = 60°.
- Construct the locus of points inside the triangle which are equidistant from BA and BC.
- Construct the locus of points inside the triangle which are equidistant from B and C.
- Mark the point P which is equidistant from AB, BC and also equidistant from B and C. Measure and record the length of PB.
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.
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 centre of a wheel of a bicycle going straight along a level road.
Describe the locus of points at distances less than 3 cm from a given point.
Describe the locus of points at distances less than or equal to 2.5 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.
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.
