Two circles of equal radii are intersecting at points of (0,1) and (0,-1). 
The tangent at the point (0,1) to one of the circles passes through the centre of the circle, then find the distance between the centers of these circles. 

Clearly, these circles intersect orthogonally. 
Let’s the centers be $C_1$ and $C_2$ and point of intersection (1,0) is P
$\Delta PC_1C_2$ is a right triangle.
Point O is the point of intersection of the line joining (0,1) and 0,-1) and the centres of the circles. 
$PO$ is 1 unit. 
Now angle $PC_1O=45$
In $\Delta PC_1O$
$$\sin 45=\frac 1r$$ where r is the radius of the circle 
$$r=\sqrt 2$$
Therefore distance between the circles is $$2r=2\sqrt 2$$ 
But the answer given is 2 unit. What is going wrong?
 A: 
Question: Two circles of equal radii are intersecting at points of $(0,1)$ and $(0,-1)$. The tangent at the point $(0,1)$ to one of the circles passes through the centre of the circle, then find the distance between the centers of these circles.

Answer: According to the question the figure is as follows

Given that the radius of the two circles are same and let it be $~r~$. 
So $\text{C$_1$B}=\text{BC$_2$}=\text{C$_2$D}=\text{C$_1$D}=r$
Now from the figure it is clear that $\text {C$_1$BC$_2$D}$ is a square and hence the diagonal $\text {C$_1$C$_2$}=\sqrt 2~r$.
Again $\text{DO}=\text{BO}=\dfrac 12~\text{DB}=\dfrac 12~\cdot 2=1~.$
Also $\text O\equiv(0,0)$, the mid-point of $\text {C$_1$C$_2$}$ and $\text{OC$_1$}=\text{OC$_2$}=\dfrac{\sqrt 2~r}{2}=\dfrac{r}{\sqrt 2}$.
Now by Pythagoras' theorem, for the triangle $~\text{C$_1$BC$_2$}~$,
$$\text{C$_1$B}^2=\text{C$_1$O}^2+\text{BO}^2$$
$$\implies r^2=1+\dfrac{r^2}{2}$$
$$\implies \dfrac{r^2}{2}=1$$
$$\implies r=\sqrt 2$$
Hence the distance between the centers of given two circles $\text {C$_1$C$_2$}=\sqrt 2\cdot\sqrt 2=2$
A: Note that the distance between the centers is not $2r$ as you have indicated in your post. 
Instead, it is $2PO = 2$, because $\triangle PC_1O$ is a 45-45 right triangle and $PO = C_1O = \frac12 C_1C_2$.
