# Area of circles: represent $x$ in terms of $r_1$ and $r_2$

See the image. Area of green and red regions are equal. Can you represent $x=|O_2D|$ in terms of $r_1$ and $r_2$ for $r_1> r_2$ ?

Edit: The point $O_1$ does not enter in the region of small circle.

• what is $x$‌‌‌? – user59671 May 23 '13 at 2:12
• We have to find $x$ :) – Berci May 23 '13 at 2:13
• It is on the figure, but you may need sharp eyes to see it :) – Lord Soth May 23 '13 at 2:13
• It is the length of that "small perpendicular piece" that starts for the center of the smaller circle and "goes down a little." – Lord Soth May 23 '13 at 2:14
• @LordSoth exactly that is. Because of the resolution of the image, it is difficult to see. – newzad May 23 '13 at 2:24

If we split the green area into two sections (P and Q) along line M, we can find each of their areas in terms of the radius and angle of one of the circles, by subtracting the area of the triangle formed by the two radii and M from the area of the whole sector. (I'm going to use big $R$ for $r_1$ and little $r$ for $r_2$.)
$$Area\space of\space P = \frac{\theta}{2\pi}{\pi{r^2}} - r^2\cos{\frac{\theta}{2}}\sin{\frac{\theta}{2}}$$ $$Area\space of\space Q = \frac{\phi}{2\pi}{\pi{R^2}} - R^2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}}$$
$$Area \space of\space Q + \space Area \space of \space P = \frac{\phi}{2\pi}{\pi{R^2}} - R^2\cos{\frac{\phi}{2}}\sin{\frac{\phi}{2}} + \frac{\theta}{2\pi}{\pi{r^2}} - r^2\cos{\frac{\theta}{2}}\sin{\frac{\theta}{2}} = \frac{1}{2}\pi {r^2}\tag{1}$$
That's four variables. We can relate $\theta$ to $\phi$ by expressing the length of $M$ in terms of each. $$2r\sin{\frac{\theta}{2}} = M = 2R\sin{\frac{\phi}{2}}$$ $$\phi = 2\arcsin{(\frac{r}{R}\cdot\sin{\frac{\theta}{2}})}\tag{2}$$ Pluging $(2)$ into $(1)$ should be sufficient to express $\theta$ in terms of $r$ and $R$.
Now, $x$ is $R$ less the length of the rhombus. $$x = R - r\cos{\frac{\theta}{2}} - R\cos{\frac{\phi}{2}}$$ and the rest is algebra.