Let $f(x) = \sqrt{1-x^2}-(px+q)$ and $\delta_0 = \frac{\sqrt{2}-1}{2} \approx 0,2071$.
With this notation, the given problem reads: $|f(x)| \leq \delta_0, \;\; x \in \left [ 0;1 \right ]$.
As seen from the expression, the problem can be interpreted graphically as a constraint
on the absolute difference between the two functions: $\sqrt{1-x^2}$ and $px + q$.
Note, that the pair $(x,y) = (x, \sqrt{1-x^2})$, for $x \in \left [ 0;1 \right ]$ is the arc of a unit circle in the 1st quadrant of the coordinate system.
Hence, a symmetry argument intuitively suggests $p = -1$ as a solution. Another way to see this, is to use the endpoint constraints:
$x = 0$: $|1-q| \leq \delta_0 \Rightarrow 1-\delta_0 \leq q \leq 1 + \delta_0\;\;\;\;\;(1)$
$x = 1$: $|-p-q| \leq \delta_0\;\;\;\;\;(2)$
Obviously, putting $p = -1$ makes the two conditions identically valid. In addition, we have the function maximum constraint:
$0 < x < 1$: $|\sqrt{1+p^2}-q| \leq \delta_0 \;\;\;\;\;(3)$
The inequality follows from two observations:
(a). $f’’(x) = \frac{-1}{(1-x^2)^{\frac{3}{2}}} \leq 0$ (i.e. $f$ is truly concave on the open interval $]0;1[$, so if an extremum is detected, it must be a maximum).
(b). $ f’(x_0) = 0 \Rightarrow x_0 = -\frac{p}{\sqrt{1+p^2}}$.
Note the minus sign (we know from $(2)$, that $p < 0$). The condition: $|f(x_0)| < \delta_0$ yields $(3)$.
The presumed solution: $p = -1$ also satisfies $(3)$, because:
\[ \left | \sqrt{1+(-1)^2}-q \right | \leq \delta_0 \Rightarrow q \geq \sqrt{2}-\delta_0 = 1+\delta_0 \Rightarrow q = 1+\delta_0\] – which is the upper limit of $q$. We can conclude, that $(p,q) = ( -1, 1+\delta_0) = \left (-1, \frac{\sqrt{2}+1}{2} \right )$ is a solution to the problem.
We need to check for other possibilities:
The conditions $(2)$ and $(3)$ imply:
$-\delta_0 -p \leq q \leq \delta_0 -p \;\;\;\;(2a)$
$\sqrt{1+p^2}-\delta_0 \leq q \leq \sqrt{1+p^2}+\delta_0 \;\;\;\; (3a)$
Note from $(3a)$ and $(1)$, that $p$ is restricted to the interval $-1 \leq p \leq 0$. We can graphically illustrate the set of possible solutions:
Only $(p,q)$-pairs in the intersection of the yellow and green sets are allowed. Thus only the o-marked corner point at $(-1,1+\delta_0)$, is a valid solution. Indeed, the lower limit in $(3a)$ intersects the upper limit in $(2a)$ in $p = -1$:
\[\sqrt{1+(-1)^2}-\delta_0 = \delta_0 - (-1) \Rightarrow \sqrt{2}-\delta_0 = 1+\delta_0\Rightarrow 2\delta_0 = \sqrt{2}-1 \Rightarrow \delta_0 = \frac{\sqrt{2}-1}{2}\;\;\;\;TRUE\]
We can conclude, that $(p,q) = \left (-1, \frac{\sqrt{2}+1}{2} \right )$ is the only solution.
