Find centre of circle $C_2$ given equation of $C_1$ and 2 points of intersection with $C_1$ Given that radius of $C_2$ is $\sqrt 2$, equation of $C_1$ is $x^2 - 4x + y^2 -8y=0$ and $C_2$ intersects $C_1$ at origin and where $x=2$, find the coordinates of centre of $C_2$ if the centre of $C_2$ is in $C_1$.
I've found that the second point of intersection is $(2, 4-2\sqrt 5)$, tried letting the coordinates of the centre be $(a, b)$, then coming up with 2 equations whereby the euclidean distance from the centre of the $C_2$ and the two points of intersection equals $\sqrt 2$.
However I end up with 2 equations with unknown variables $a, b$ to the power of 2 and am unsure of how to solve from here on.
Any help will be greatly appreciated thank you.
 A: The center of those circles that cross the origin and whose radius is $\sqrt 2$ sit on a circle of radius $\sqrt 2$ centered at the origin. So our circles are described by$$(x-a)^2+(y-b)^2=2.$$ For their centers we have $$a^2+b^2=2\ \ \Rightarrow  b=\sqrt{2-a^2}$$
so, finally, the equations of our circles are $$(x-a)^2+(y-\sqrt{2-a^2})^2=2.$$ I consciously omitted the minus sign above because we are interested only in those circles whose centers are located in the first quadrant. The following figure depicts the two extremes:

Intuition tells that there exists a little circle crossing the big one at a point whose $x=2$. The center of the required circle is centered on the goose poop colored circle somewhere between the two black dots (in the first quadrant.)

The $y$ coordinate of the point on the large circle is $$4-\sqrt{20}$$ if $x=2$. I consciously omitted the $+$ sign because our $y$ is negative.
Now, the question is: which point of the goose poop colored central circle is $\sqrt 2$ units away from the point $(2,4-\sqrt{20})$?
In the first quadrant the coordinates of the points on the central circle are $(a,\sqrt{2-a^2})$. The square of the distance of such a point from $(2,4-\sqrt{20})$ is
$(a-2)^2+(\sqrt{2-a^2}-(4-\sqrt{20}))^2$ then the equation one will have to solve is
$$(a-2)^2+(\sqrt{2-a^2}-(4-\sqrt{20}))^2=2.$$
Alpha solved the equation above for me and the solution turned out to be equal to the solution given by the OP in a comment to Michael's answer.
A: If  "the center of $C_2$ is in $C_1$" it means "the centre of $C_2$ is placed on  $C_1$" then we have the following hint.
For the $C_2(a,b)$ we have the following system.
$$a^2+b^2=2$$ and
$$a^2+b^2-4a-8b=0.$$
Now, check that the distance between $(a,b)$ and $(2,4\pm2\sqrt5)$ is equal to $\sqrt2$, which gives that $C_2$ does not exist.
If it means "the center of $C_2$ is inside of $C_1$" then the answer indeed is very ugly. 
