freemind wrote: In the usual coordinate system $ xOy$ a line $ d$ intersect the circles $ C_1:$ $ (x + 1)^2 + y^2 = 1$ and $ C_2:$ $ (x - 2)^2 + y^2 = 4$ in the points $ A,B,C$ and $ D$ (in this order). It is known that $ A\left( - \frac32,\frac {\sqrt3}2\right)$ and $ \angle{BOC} = 60^{\circ}$. All the $ Oy$ coordinates of these $ 4$ points are positive. Find the slope of $ d$. Hello Andrew, here is my solution. Firstly one can show easilly that $ \angle{KLA} = 60^o$ and therefore $ \angle{OBC} = 60^o\Longrightarrow \angle{OCB} = 60^o$ and finally $ \angle{OFD} = 60^o$. Now the triangle $ GDF$ is isosceles and $ \angle{GFD} = 60^o$, which means that the triangle $ GFD$ is equilateral. Therefore the abscissa of point $ D$ must be equal to $ OM$ where $ M$ the midpoint of the segment $ GF$, so $ D = (3,\sqrt {3})$ and now $ d = \frac {DT}{AT} = \frac {\sqrt {3}}{9}$. I hope that my calculations are ok, but forgive me if not. Nick Rapanos

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