MHB Find the radius of the small circle O_2

[Albert1]

find the radius of the small circle O_2:
View attachment 1660

 

[I like Serena]

Re: find the radius of the small circle O_2

Spoiler

We can draw a triangle from the left bottom corner, to the center of the small circle, to the right bottom corner.
And then split it into 2 rectangular triangles.

Let's call the radius of the small circle x.
Then the left rectangular triangle has hypotenuse (36-x) and horizontal side at the x-axis (x).
And the right rectangular triangle has hypotenuse (36+x) and horizontal side at the x-axis (36-x).

Since they share their third side, the following equation must hold (Pythagoras):
$$(36-x)^2 - x^2 = (36+x)^2 - (36-x)^2$$
$$(36+x)^2 - 2(36-x)^2 + x^2 = 0$$
$$(36^2+2\cdot 36 x +x^2) - 2(36^2-2\cdot 36 x + x^2) + x^2 = 0$$
$$6\cdot 36 x = 36^2$$
$$x = 6$$
$\blacksquare$

 

[Opalg]

Re: find the radius of the small circle O_2

find the radius of the small circle O_2:
https://www.physicsforums.com/attachments/1660

[sp]

View attachment 1662​

Let $r$ be the radius of the green circle. Draw the triangle whose vertices are the two ends of the red baseline and the centre of the green circle (the points labelled O, O_2 and O-1 in the diagram). The lengths of its sides are $36-r$, $36+r$ and $36$. The angle labelled $\theta$ has $\cos\theta = \dfrac r{36-r}$. The cosine rule then gives the equation $(36+r)^2 = 36^2 + (36-r)^2 -2\cdot36(36-r)\dfrac r{36-r},$ which simplifies to $r=6.$[/sp]

Edit. Yet again, ILS got there first (and I never even noticed).

 

[I like Serena]

Re: find the radius of the small circle O_2

Edit. Yet again, ILS got there first (and I never even noticed).

Must be my new avatar.
Sometimes it makes me feel imaginary.

 

[alyafey22]

Re: find the radius of the small circle O_2

Must be my new avatar.
Sometimes it makes me feel imaginary.

The invisible : not recognizable neither by name nor by avatar living in his imaginary complex paradigm , just kidding (Punch) .