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Two circles with radii r1 and r2 touch each other externally. Let r be the radius of circle which touches these two circles as well as the common tangent to the two circles.Prove that: 1/sq root r=1/sq root r1 + 1/sq root r2

Two circles with radii r1 and r2 touch each other externally. Let r be the radius of circle which touches these two circles as well as the common tangent to the two circles.Prove that: 1/sq root r=1/sq root r1 + 1/sq root r2

Grade:10

1 Answers

Arun
25768 Points
3 years ago
 

the main idea is to find the distance b/w the foot of perpendiculars from the center of the two given circles to the common tangent by two ways and then equating the two equations obtained to get the value of radius.

 

1st equation

 

form a right triangle by taking the distance b/w the centers of the circles as hipotenuse and solve for the base using pythagores th.

 

2nd equation

 

form right angled triangles (2) by joining the centre of the required circle with the centres of the two given circles and solve for the base and then sum up the two distances and then equate it with the distance obtainesd in the first equation

 

r  =  r1 * r2/( sq. root r1+ sq. root r2)2

1/sq root r = (sq. root r1+ sq. root r2)/ (sq root (r1*r2))

we see that

1/sq root r = 1/sq root r1  +  1/sq root r2

 

hope it helps

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