How prove this $|ON|\le \sqrt{a^2+b^2}$ let ellipse $M:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$,and there two point $A,B$ on $\partial M$,and the point $C\in AB$ ,such $AC=BC$,and the Circle $C$ is directly for the AB circle,for any point $N$ on $\partial C$,
show that
$$|ON|\le\sqrt{a^2+b^2}$$

my try: let 
$$A(x_{1},y_{1}),B(x_{2},y_{2}),C(\dfrac{x_{1}+x_{2}}{2},\dfrac{y_{1}+y_{2}}{2})$$
then
$$\dfrac{x^2_{1}}{a^2}+\dfrac{y^2_{1}}{b^2}=1,\dfrac{x^2_{2}}{a^2}+\dfrac{y^2_{2}}{b^2}=1$$
so the circle $C$ equation is 
$$\left(x-\dfrac{x_{1}+x_{2}}{2}\right)^2+\left(y-\dfrac{y_{1}+y_{2}}{2}\right)^2=r^2$$
where
$$4r^2=(x_{1}-x_{2})^2+(y_{1}-y_{2})^2$$
let $N(c,d)$,then
$$\left(c-\dfrac{x_{1}+x_{2}}{2}\right)^2+\left(d-\dfrac{y_{1}+y_{2}}{2}\right)^2=r^2$$
How prove
$$c^2+d^2\le a^2+b^2?$$
Thank you 
 A: Following Tian's observation and taking
$$A:(a\cos\theta,b\sin\theta),$$
$$B:(a\cos\phi,b\sin\phi)$$
we have to prove
$$4\|OC\|^2+8\|OC\|\|AC\|+4\|AC\|^2\leq 4a^2+4b^2,\tag{1} $$
or:
$$a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2+a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2+2\sqrt{\left(a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2\right)\cdot\left(a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2\right)}\leq 4a^2+4b^2,\tag{2}$$
equivalent to:
$$\sqrt{\left(a^2(\cos\theta+\cos\phi)^2+b^2(\sin\theta+\sin\phi)^2\right)\cdot\left(a^2(\cos\theta-\cos\phi)^2+b^2(\sin\theta-\sin\phi)^2\right)}\leq a^2(\sin^2\theta+\sin^2\phi)+b^2(\cos^2\theta+\cos^2\phi),\tag{3}$$
and to:
$$a^4\left(\sin^2\theta-\sin^2\phi\right)^2+b^4\left(\cos^2\theta-\cos^2\phi\right)^2+a^2b^2\left(\sin(2\theta)-\sin(2\phi)\right)^2\leq a^4(\sin^2\theta+\sin^2\phi)^2+b^4(\cos^2\theta+\cos^2\phi)^2+2a^2b^2(\sin^2\theta+\sin^2\phi)(\cos^2\theta+\cos^2\phi),\tag{4}$$
that further reduces to:
$$4a^2 b^2(\sin\theta\cos\theta-\sin\phi\cos\phi)^2\leq 4a^4\sin^2\theta\sin^2\phi+4b^4\cos^2\theta\cos^2\phi+2a^2b^2(\sin^2\theta\cos^2\theta+\sin^2\phi\cos^2\phi+\sin^2\theta\cos^2\phi+\sin^2\phi\cos^2\theta),\tag{5} $$
and finally to:
$$2a^2 b^2(\sin^2\theta-\sin^2\phi)(\cos^2\theta-\cos^2\phi)\leq 4(a^2\sin\theta\sin\phi+b^2\cos\theta\cos\phi)^2\tag{6} $$
that is trivial since the LHS is non-positive while the RHS is non-negative.
A: Jack has already written quite an answer to this problem, but I couldn't believe that it didn't have more elegant solution. As my sense of beauty didn't leave me at peace, I couldn't help but find some geometrical ideas behind this problem. I must also add that I really enjoyed solving it.
As Tian has already noted, it suffices to show that $OC + CA \leqslant \sqrt{a^2 + b^2}$.
Let's reflect points $A$, $B$ and $C$ across the point $O$. We get a parallelogram $ABA'B'$ and the desired sum $\color{green}{OC + CA}$ equals to $\frac{1}{4}$ of the perimeter of $ABA'B'$ (since $CC'$ is the middle line of $ABA'B'$).

So we want to show that perimeter of parallelogram inscribed in an ellipse is not bigger than $4\sqrt{a^2 + b^2}$.
In other words, it would be great if we knew how to maximize the perimeter of such a parallelogram. Fortunately, humanity knows how to solve this problem, namely:

Among all the parallelograms inscribed in a given ellipse
  those of maximal perimeter have the property that any point on the ellipse is the vertex of exactly one.

(paraphrasing result of this article)
Moreover, according to Lemma from that article, the parallelogram has maximal perimeter iff tangent lines at adjacent vertices are perpendicular to each other.

This leads us to the final fact that the maximum perimeter in originally stated problem is $4\sqrt{a^2 + b^2}$ because choosing point $A$ so that it lies on the coordinate axis produces exactly rhombus with a side equal to $\sqrt{a^2 + b^2}$.
