# Need help here!

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 $$OO^{'}$$. . Note by Karthik Venkata
5 years, 3 months ago

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- 5 years, 3 months ago

This is clear by symmetry.

- 5 years, 3 months ago

This 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 $M_1=M_2=M$.

- 5 years, 3 months ago

Thank You !

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago

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

- 5 years, 3 months ago