Given \( 2 \) circles with centres \( O \) and \( O^{'} \) in the Euclidean plane, one draws a couple of transverse common tangents to these circles, \( T_{1} \) and \( T_{2} \), and mark their intersection point as \( M \). Prove that \( M \) lies on \( OO^{'} \).

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TopNewestThis is essentially homothety, consider the following proof:

Suppose the tangents intersect \(OO'\) at \(M_1,M_2\). Prove that \(\frac {OM_1}{O'M_1}=\frac {OM_2}{O'M_2}\) and that this is sufficient to conclude that \(M_1=M_2=M\). – Xuming Liang · 2 years, 1 month ago

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– Karthik Venkata · 2 years, 1 month ago

Thank You !Log in to reply

– Aditya Kumar · 2 years, 1 month ago

You could also bash the problem and prove that M lies on the equation of OO'.Log in to reply

– Karthik Venkata · 2 years, 1 month ago

Yeah, but I don't like it. I only like Euclid-style proofs, they are more beautiful.Log in to reply

– Aditya Kumar · 2 years, 1 month ago

Yes u r right euclid style proofs r really awsome. but they require high level intuition. bashing doesn't require much intuition.Log in to reply

@Xuming Liang @Nihar Mahajan @Daniel Liu – Karthik Venkata · 2 years, 1 month ago

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– Daniel Liu · 2 years, 1 month ago

This is clear by symmetry.Log in to reply