# 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.

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.

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.

• I believe it means the latter- "center of $C_2$ is inside of $C_1$". I have the answer key- it is supposed to be $(1.22, 0.710)$, but I can't get another equation to solve for both $a$ and $b$. – user445310 Dec 16 '17 at 8:49
• @Michael Rozenberg: Yes, the circle in question exists. I calculated the coordinates of its center. – zoli Dec 16 '17 at 9:35
• I am not ready for this ugly work. – Michael Rozenberg Dec 16 '17 at 9:47