Finding all $a,b,c$ such that $(ax)^2+(b+cy)^2 \leq 1$ whenever $x^2+y^2 \leq 1$. Suppose $x,y,a,b,c \in \mathbb R$ and suppose that $$(ax)^2+(b+cy)^2 \leq 1$$ for all $x,y$ such that $x^2+y^2 \leq 1$. What are the allowed values of $a,b,c$? 
Plugging in $x=1$, $y=0$ we get $a^2+b^2 \leq 1$. Similarly for $x=0,y=1$ we get $(b+y)^2 \leq 1$. Are there any additional constraints imposed on $a,b,c$?
 A: One is a circle centered at $(0,0)$ and with radius $1$.
The other is an ellipse
$$
\left( {ax} \right)^{\,2}  + \left( {b + cy} \right)^{\,2}  = \left( {{x \over {1/a}}} \right)^{\,2}  + \left( {{{b/c + y} \over {1/c}}} \right)^{\,2}  \le 1
$$
centered at $(0, -b/c)$, and with semi-axes $(1/a, 1/c)$.
You want (if I understood exactly) that the ellipse be contained in the circle.
Can you proceed from here , using , e.g., the polar coordinates ?
Otherwise please specify what tools bag you have.
A: I played around for a bit
and came to a point
where I didn't want 
to go further.
I'll show what I did
in the hope that
someone else
might find this useful.
Assume that
$x$ and $y$
are on the boundary,
so
$x^2+y^2 = 1
$.
This becomes
$\begin{array}\\
1
&\ge a^2x^2+b^2+2bcy+c^2y^2\\
&=a^2(1-y^2)+b^2+2bcy+c^2y^2\\
&=a^2+b^2+2bcy+(c^2-a^2)y^2\\
\end{array}
$
or
$1-a^2-b^2
\ge 2bcy+(c^2-a^2)y^2
$.
The equality case for this is
$\begin{array}\\
y
&=\dfrac{-2bc\pm\sqrt{4b^2c^2+4(c^2-a^2)(1-a^2-b^2)}}{2(c^2-a^2)}\\
&=\dfrac{-bc\pm\sqrt{b^2c^2+(c^2-a^2)(1-a^2-b^2)}}{(c^2-a^2)}\\
&=\dfrac{-bc\pm\sqrt{a^4 + a^2 b^2 - a^2 c^2 - a^2 + c^2}}{(c^2-a^2)}\\
\end{array}
$
And that's where
I stop.
