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# Circles with common tangent

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.

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Note by Agr Amul
2 years, 7 months ago

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Drop 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. · 2 years, 7 months ago