# 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
2 years, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

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 $$M_1=M_2=M$$.

- 2 years, 10 months ago

Thank You !

- 2 years, 10 months ago

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

- 2 years, 10 months ago

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

- 2 years, 10 months ago

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

- 2 years, 10 months ago

- 2 years, 10 months ago

This is clear by symmetry.

- 2 years, 10 months ago