Let's say we have the curves $y=x^{p^{\pm1}}$ on $[0,1]$ for $p>1$, which are symmetric about $y=x$ the case $p=1$. Let $Q=(q,q)$ ($0 < q < 1$) be the center of the desired circle $C$. Let $r$ be the minimum distance from Q to the curves. Let $P=(x,y=x^p)$ be a closest point to $Q$ (fix $y=x^p$). Then $C$ is tangent to $x^p$ at $P$ (proof for $p\in\mathbb{N}$: Lemma 2, p. 330 in Jacobson, Basic Algebra 1, 2nd ed.). This amounts to an equation of slopes. The slope, $\beta$, of $x^p$ at $P$ is found from the derivative. The slope of the tangent to the circle at $P$ is the negative reciprocal (perpendicular) slope of the radial line $\overline{QP}$. Thus $ \frac{x^p-q}{x-q}=-\frac{1}{px^{p-1}}=-\beta^{-1} \tag{1a} $ implicitly relates $x$ and $q$ for a given $p$, which can be written: $ px^{p-1}(x^p-q)+(x-q)=0 \tag{1b} $ $ q=\frac{x(1+px^{2p-2})}{1+px^{p-1}} \tag{1c} $ $ q'=\frac{p^3x^{3(p-1)}+p(2p-1)x^{2(p-1)}-p(p-2)x^{p-1}+1}{p^2x^{2(p-1)}+2px^{p-1}+1} =\frac{\beta^3+(2-\frac1p)\beta^2-(p-2)\beta+1}{(\beta+1)^2} ~\tag{1d} $ Now the distance $r$ from $P$ to $Q$ has square $ r^2=(x^p-q)^2+(x-q)^2 \tag{2} $ and we wish to maximize (2) on the interval $(0,1)$ subject to the constraint (1) and to the constraint that the resulting circle given by $q$ and $r$ never crosses the dual power curves elsewhere. However, in order for the $r$ to be maximal, (I think) this means that each $x$ at which $C$ meets $x^p$ must be a local or global maximum of $r$. In fact, there may be either one or two such points; for a given $p>1$, $q$ and $r$ are uniquely determined, but $P$ will have one or two solutions, depending upon whether $p$ is past a certain threshold, around $5.2$.
Here is an animated graph of the solutions for $p$ varying from $1$ to $10$. The solution circle $C$ is shown in green. The dotted blue & purple (cyan/violet) circles show perturbations of $q$. The dashed grey lines show the critical points of $q$, or roots of $q'$. Between these two lines is the area where $q$ displays "retrograde motion" as $x$ increases. The purple and blue curves show how $q$ and $r$ depend on $x$. The green radial line represents $\overline{QP}$, or an arbitrary choice of one of the two possibilities for $p>5.2$.

I used WNY's suggestion to find the solution, but had to first overcome a numerical instability issue. The red arcs display crossover outside the power curves, which results from numerical inaccuracies. To overcome these, I used symmetry properties and the inequalities $0 < r < \min(r,1-q)$ and $y < q < x < 1$. If there is interest, I could post an animation or sage code demonstrating this as well. When time permits, I will update and revise the below with some further observations, and hopefully find an analytic value for the threshold value of $p$ around $5.2$, unless someone beats me to it (or knows where the problem is already solved)!

Let $f(x)=\frac12r^2$ with $r$ as in (2) and $q$ as in (1). Its local extrema occur when its derivative vanishes. $ \eqalign{f\,'(x) &=(x^p-q)(px^{p-1}-q')+(x-q)(1-q') \\&=\left(px^{p-1}(x^p-q)+(x-q)\right)-q'\big(x^p+x-2q\big) \tag{3a} } $ Since the circle is tangent to the curve at all local extrema, we can use (1b) to eliminate the first parenthesized term, $ f\;'(x)=r\;r'=-q'\cdot\big(x^p+x-2q\big) \tag{3b=1$\wedge$3a} $ $ \matrix{ r' &=& \frac{2}{r} \left(q-\frac{x+y}{2}\right) &\cdot& q' \cr \frac{dr}{dx} &=& \frac{dr}{dq} &\cdot& \frac{dq}{dx} } \tag{3c} $ resulting in two factors, where $q'$ is given in (1d). Thus, $f$ & $r$ have the same critical points as $q$, plus one more, representing the difference in coordinate values between two points on $y=x$, namely, between $Q=(q,q)$ and the midpoint of $P=(x,y)$ and its "dual" point $(y,x)$, with coordinate $\frac{x+y}{2}$. Let's assume for now that $q'\ne0$. This implies $x^p+x-2q=0$, i.e., $q=\frac{x^p+x}{2}=\frac{x+y}{2}$. With the slope equation (1b) above, then, we have $ \eqalign{ 0 &=px^{p-1}\left(x^p-\frac{x^p+x}{2}\right)+\left(x-\frac{x^p+x}{2}\right) \\&=\left(px^{p-1}-1\right)\left(\frac{x^p-x}{2}\right) \tag{3.1a} } $ which has one real root in $(0,1)$, at $x=p^{-1/(p-1)}=p^{1/(1-p)}$. (The other roots are all $(p-1)^\text{th}$ roots of $1$ and $\frac1p$, which, aside from a negative root of each for odd $p$, lie off the real axis). Thus $ x=p^{1/(1-p)} \qquad y=x^p=p^{p/(1-p)} \qquad q=\frac{x+y}{2}, \tag{3.1b} $ the distance and area are $ r=\frac{|x^p-x|}{\sqrt{2}} \qquad A=\frac{\pi}{2}|x^p-x|^2, \tag{3.1c} $ and $ y-q=x^p-q=\frac{x^p-x}{2}=\frac{y-x}{2}=-(x-q) \,. \tag{3.1d} $ In fact, (3.1) gives the global maximum of $f$ for $1. Using L'Hopital's rule as $p\rightarrow\infty$, $ \eqalign{ &\log x= -\frac{\log p}{p-1} \rightarrow -\frac1p \rightarrow 0, \quad \log y= -\frac{\log p}{1-\frac1p} \rightarrow -\infty \\&\implies \quad x\rightarrow1, \quad y\rightarrow0, \quad q\rightarrow\frac12, \text{ and } \quad r\rightarrow\frac{1}{\sqrt2} \,. \tag{3.1e} } $ Similar analysis show that $x\rightarrow y\rightarrow q\rightarrow\frac1e$ as $p\rightarrow 1$. Equations (3.1a-d) must maximize $f$ for all $p$ for which $q'\ne0$ on $(0,1)$. The limiting behavior, however, shows that this cannot hold for $p$ sufficiently large. This is because the circle meets $(x,y=x^p)$ at a tangent, but must also have a smaller radius of curvature to stay above the curve. This rules out the limit point at $(1,0)$ as $p\rightarrow\infty$.
The problem is that, for $p$ large enough (the threshold is around $5.72$), the other term, $q'$, also has real roots in $(0,1)$, and the minimum crosses over to one of these; the root of $x+x^p-2q$ becomes a maximum at the crossover. The graph below shows the approximate threshold case. The extrema of the green circle's area occur when the product (i.e. either) of the blue ($2q-x-x^p$) and/or red-brown ($q'$) curves has a root. For $1 < p < 5.72$, $q'$ is strictly positive on $(0,1)$, while for $p$ above the threshold, two new roots are introduced.

We need to ensure that we are always choosing the solution with minimal $r$. We must choose the least positive root $x\in(0,1)$ of $q'(x)$, when it exists.
To check these statements, we can take the second derivative of $f$: $ \eqalign{ f\;''(x)&= -q''\left(x^p+x-2q\right) -q'\left(px^{p-1}+1-2q'\right)\\&= 2(q')^2 -q''\left(x^p+x-2q\right) -q'\left(px^{p-1}+1\right) \tag{4} } $ Its sign tells us whether each critical point is a local minimum or maximum. Or, as already mentioned, we can check that the radius $r$ is less than the radius of curvature, $R$, of $y=x^p$ at $P$, given by $ \eqalign{ R^2 &= \frac{\left(1+(y')^2\right)^3}{(y'')^2} &= \frac{\left(1+(px^{p-1})^2\right)^3}{\left(p(p-1)x^{p-2}\right)^2} &= \frac{\left(1+p^2x^{2(p-1)}\right)^3}{p^2(p-1)^2x^{2(p-2)}} } \tag{5} $ which should lead us to the cutoff we desire when we equate $R^2$ with $r^2$ above (if it is not equivalent to $q'=0$).
Here is an animation of the result of using (3.1) for $p\in[1,11)$ with increments of $\frac1{10}$. You can clearly see how the root becomes a local maximum after the crossover threshold.

Before evaluating critical points using (4) or (5), we will also need to find the critical points of $q$, using the numerator of $q'$ from (1d): $ 0=p^3x^{3(p-1)}+p(2p-1)x^{2(p-1)}-p(p-2)x^{p-1}+1 \tag{3.2a} $ which I call equation (3.2) since it represents a second branch of (3) corresponding to the first, previously discounted, factor of (3b). With $\beta=px^{p-1}\in(0,p)$ (the common slope at $P$), we have the cubic $ 0=\beta^3+(2-\tfrac1p)\beta^2-(p-2)\beta+1 \tag{3.2b} $ which for $p$ sufficiently large has roots near $ px^{p-1} = \beta \approx \frac1{p-2},~ -1\pm\sqrt{p-1-\frac1{p}}+\frac1{2p}, \tag{3.2$\approx$} $ one of which is in the interval $(0,1)$ and must therefore be our solution. Its precise location can be expressed as (I suspect) $ \beta = \gamma - \frac{2p-1}{3p} + \frac{3p^3-2p^2-4p+1}{9p^2\gamma} \tag{3.2$\beta$} $ for $\gamma$ given by (the cube root of) $ \gamma^3 = \left(p + 1\right) \sqrt{\frac{-p^4+7p^3-11p^2+7p-1}{27p^3}} - \frac{9p^4-p^3-3p^2+6p-1}{27p^3}. \tag{3.2$\gamma$} $ Interestingly, the complex roots of (3.2a) for integral $p>2$ all seem to lie in the annulus $\frac12<|x|<1$.
To be continued...