Request a call back

Transverse common tangent
Asked by SSHHII | 13 Oct, 2009, 10:59: PM Expert Answer

If we form the triangle having sides as the line joining centres of the circle of length d, radius of the bigger circle from the point of tangency and the line parallel  to the transverse tangent joining centre of smaller circle and perpendicular on the radius of the biggger circle. This line will be equal in length to the tranverse tangent let it be X.

The sides will be d, R+r, the length of tangent.

also this triangle forms a right angled triangle whose hypotnuse is d.

Thus applying Pythagoras theorem

d2=(R+r)2+X2

X2=d2-(R+r)2

Thus length of the transverse tangent = d2-(R+r)2

d2=(R+r)2+X2

Answered by | 21 Oct, 2009, 06:02: PM
CBSE 10 - Maths
Asked by tejasdd | 28 May, 2010, 08:18: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by ketanvai | 23 Apr, 2010, 06:35: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by srprusty | 10 Apr, 2010, 05:43: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by s.sreeram | 11 Mar, 2010, 08:14: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by | 10 Mar, 2010, 07:26: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by rinki94 | 10 Mar, 2010, 03:59: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:24: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:08: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by BhavSimran | 10 Mar, 2010, 12:05: PM ANSWERED BY EXPERT
CBSE 10 - Maths
Asked by nishac | 10 Mar, 2010, 09:12: AM ANSWERED BY EXPERT