What's the meaure of the radius in the circle below?

Question

For reference: Calculate r in the figure (O is the center) R = 4 (Answer:$\sqrt{\frac{8}{3}}$)

My progress:

$\triangle AOT:\\ AT^2 = 4^2+4^2 \implies AT = 4\sqrt2$

Th Apollonius:

$\triangle AEB:\\ BE^2 +(4+r)^2 =2OE^2 + \frac{8^2}{2} \implies \\ BE^2+16+8r+r^2=2OE^2 +32\\ BE^2+2OE^2+8r+r^2 -16=0$

...???

Answer 1

From point $ \small C$, drop perp $ \small CE$ to $ \small OD$ and $ \small CH$ to $ \small AB$.

$ \small OH = CE = \sqrt{OC^2 - OE^2} ~$.
Note $ \small OC = R-r, OE = R/2$ (as $OC = CD = R-r$).

Once you find $ \small OH$,

$ \small AC^2 = (AO+OH)^2 + CH^2$
$ \small \implies 2 AO \cdot OH = AC^2 - CH^2 - AO^2 - OH^2$.

Note $ \small CH = OE = R/2, AC = R + r, AO = R$

It does simplify quickly and you should get the relationship $R^2 = 6r^2$

Answer 2

Let $ O = (0, 0)$, $ E = (x,y)$ .

We have

• $AE^2 = (4+r)^2 = (4+x)^2 + (0+y)^2 $
• $OE^2 = (4-r)^2 = (0+x)^2 + (0+y)^2 $

By symmetry, one can observe that $ y = \frac{4}{2} = 2$.
Note: If you don't want the symmetry argument, then use
$TE^2 = (4-r)^2 = (0+x)^2 + (4 - y)^2 $ from which we get $ y = 2$.

Solve this system (EG Show $ x = 2r -2 $, and substitute back) to show that $ r = \sqrt{8/3}$.