# Two circles with radii 'a' and 'b' touch each other at a point. A common direct common tangent is drawn to both the circles. Another small circle is nested between the two circles and the tangent...

Two circles with radii 'a' and 'b' touch each other at a point. A common direct common tangent is drawn to both the circles. Another small circle is nested between the two circles and the tangent such that it just touches the circles and the tangent. The radius of the smaller is c. Prove that 1/sqrt(c) = 1/sqrt(a) + 1/sqrt(b)

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### 2 Answers

Do the points M and N lie on the common tangent / lie on any other point on the circle ?

Join cntres of the cicle say C(P,a), C(Q,b) and C(O,c).Let b<a.Let M and N are on the circle C(P,a) and C(Q,b) through which direct common tangent passes through. Draw perpendiculars from O to PM and PN say OL and OK.Also draw perpendicular from Q to PM say QT.

From this we have

MN=QT=OK

OT=b-a

OP=a+c

OQ=b+c

PL=a-c

QK=b-c Thus

`OL=sqrt((a+c)^2-(a-c)^2)`

`OL=sqrt(4ac)`

`OK=sqrt((b+c)^2-(b-c)^2)=sqrt(4bc)`

Thus QT=OL+OK=`sqrt(4ac)+sqrt(4bc)`

In trianle PQT

QT=`sqrt((a+b)^2-(a-b)^2)=sqrt(4ab)`

Thus

`sqrt(4ac)+sqrt(4bc)=sqrt(4ab)`

divide by `sqrt(4abc)` ,we have

`1/sqrt(c)=1/sqrt(a)+1/sqrt(b)`