Two circles of radii 5 cm and 3 cm touch each other at A and also touch a line at B and C. The
distance BC in cms is?
plz explain with the help of diagram.
A. sqrt60
B. sqrt62
C. sqrt68
D. sqrt64
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thanks diwakar prajapati.But I know this rule co'*co'+cp*cp=o'p*o'p
Let the larger circle have center O and smaller circle have center at O'.
They touch at A.
Larger circle touch the line at B and the smaller circle at C.
Let the given line and the line passing through the centers intersect each other at P
Triangle BOP and triangle CO'P are similar, so
BO/CO'=5/3=OP/O'P=(OA+AO'+O'P)/O'P (using properties of similarity)
(OA+AO'+O'P)/O'P=5/3
(5+3+O'P)/O'P=5/3
solving we get, O'P=12
in triangle CO'P
CO'*CO'+ CP*CP=O'P*O"P
3*3+CP*CP=12*12
solving we get
CP=sqrt135
again
BP/CP=5/3 (Similarity)
(BC+CP)/CP=5/3
(BC+sqrt135)/sqrt135=5/3
solving BC=sqrt60