Tough geometry problem

The radius of the LH circle is $D+d$. Consdering the isosceles triangle with angle $\theta$ and base $r$, we see that $r \; = \; 2(D+d)\sin\tfrac12\theta$ The other two angles in this isosceles triangle are $90^\circ-\tfrac12\theta$ and so, looking at the small right-angled triangle on the right, $\sin\tfrac12\theta \; = \; \cos\big(90^\circ-\tfrac12\theta) \; = \; \frac{d}{r}$ Putting these results together we deduce that $r^2 \; = \; 2(D+d)d$

Hi, This can be done using trigonometry

$cos(\theta) = \frac{D}{D + d}$ --------- (1)

Joining the point of intersection of the two circles at the top to the center of the small circle, a beautiful triangle would be formed with,

$sin(\theta) = \frac{r}{D+d}$ --------- (2)

Squaring and adding (1) and (2), we get

$D^2 + r^2 = (D + d)^2$

let the radius of the bigger circle be R.

let the perpendicular side of the right angle triangle be x. (having angle $\theta$)

now, join the center of the smaller circle and the point of intersection between the smaller and bigger circle ( we have two point of intersections, but we will take the upper one), so now we have an another right angle triangle so, applying pythogorus theoram in our new right angle triangle :
$x^{2}+d^{2}=r^{2}$

so, $x^{2} = r^{2}-d^{2}$___ [1]

now coming back to our bigger right angle triangle

using pythogorus theoram again

we get

$x^{2}+D^{2}=R^{2}$

putting value of x^{2} from eq. 1 and value of R (since R = D+d) we have

$r^{2}-d^{2}+D^{2} = (D+d)^{2}$

$r^{2}-d^{2}+D^{2} = D^{2}+d^{2}+2Dd$

$r^{2}-2d^{2} = 2Dd$

$D= \frac {1}{2} \frac {r^{2}-2d^{2}}{d}$ ---- answer :)

(r^2)/2 = Dd + d^2

Is the line joining the centre of the bigger circle and the point of intersection of the two circles tangent to the second circle? Can it be proved?

Firstly we are assuming that the line that we can call $R$ which is the radius of the larger circle is at a tangent to the perimeter of the smaller circle and a normal to the perimeter of the larger circle (only reason I mention this is because if you zoom on the picture this is perfectly correct) and lets call this point $P$. However assuming these, we can draw a triangle from the center of the small circle along the length $D$ and then a $90°$ angle up to the point $P$ which we call length $x$, hence the hypotenuse of this triangle is $R$. We can then make another triangle which starts from the center of the smaller triangle along the length $d$ and then make another $90°$ angle up to the point $P$, the hypotenuse of this triangle is $r$. Further more, we know that $R=D+d$, using this information we can model a function of $D(r,d)$:

Using Pythagoras's theorem:

$R^{2}=D^{2}+x^{2}$

$r^{2}=d^{2}+x^{2}$

Hence, when we cancel $x$:

$R^{2}-D^{2}=r^{2}-d^{2}$ where $R=D+d$

$(D+d)^{2}-D^{2}=r^{2}-d^{2}$

Which simplifies to, $r^{2}=2d(D+d)$

Hence, $D(r,d)=\frac{r^{2}}{2d}-d$

wait, is D the radius of the larger circle???

https://www.facebook.com/groups/572200892829750/ .... link of this solution check it out final relation is r=2 * (D+d) * Cos (theta/2)

This is solved easily using pythagoras famous equation. $c^2=a^2 + b^2$

the little triangle on the right tells us: $r^2 = d^2 + h^2$

the left side triangle shares the same 'h' side, and we can also say that: $(D+d)^2 = D^2 + h^2$

Substitute 'h' obtained from the first equation into the second one, and solve fo D in terms of d and r:

$(D+d)^2 = D^2 + r^2 - d^2$

$D^2 + 2dD + d^2 = r^2 - d^2 + D^2$

$D = (r^2/2d) - d$

Not so tough, but very fun.

r/tan@ -d

use trigonometry. $tan\theta=\frac{r}{D}$

$\tan\theta\times D=r$

or,

$\tan(90-\theta)\times r=D$ You have more than enough information in this problem.

D = (r - d sin (teta) ) / (sin (teta) ) ????