Here 2 circles have a common tangent. We have to prove \( AB^{2} \)=4xy where x=AO1 and y=BO2 and O1 and O2 are centres of the circles.

Help me!

Here 2 circles have a common tangent. We have to prove \( AB^{2} \)=4xy where x=AO1 and y=BO2 and O1 and O2 are centres of the circles.

Help me!

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TopNewestDrop a perpendicular from \(O_2 \) to the segment \(AO_1\) such that it meets it at \(M\). Therefore \( |MO_1| = x - y \). Consider the right angled triangle \( \Delta O_1 O_2 M \) and apply Pythagoras theorem to obtain \( |MO_2|^2 = (x+y)^2 - (x-y)^2 \Rightarrow |AB|^2 = |MO_2|^2 = 4xy \). Hence, proved.

(\( |AB| = |MO_2| \) since the quadrilateral \( |ABO_2 M | \) is a rectangle. )

Hope this helps. – Sudeep Salgia · 1 year, 10 months ago

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– Agr Amul · 1 year, 10 months ago

But friend, How can we say O1O2 is x+y.Log in to reply