Owner of a squished heart

You can generate a heart shape by drawing a unit square (i.e., a square with side length $1$), and then attaching semicircles (each with radius $1/2$) to adjacent edges, as shown in the diagram below:

What is the radius of the smallest circle that contains this heart shape?

Let's define our frame of reference. Let's assume that the center of this unit square is the origin and that as in the figure above, the sides of the square make angle of $\frac{\pi}{4}$ with respect to the $x$- and $y$-axes. In particular, the lowest corner of the unit square is located at $(0, -\frac{\sqrt{2}}{2})$. The two semi-circles added on the upper sides of the square are centered at $(\pm\frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4})$. From symmetry, we will focus on the one in the first quadrant and parameterize its boundary as $$\left( \frac{\sqrt{2}}{4} + \frac{1}{2} \cos \left( \theta - \frac{\pi}{4} \right), \frac{\sqrt{2}}{4} + \frac{1}{2} \sin \left(\theta - \frac{\pi}{4}\right) \right),$$ for $0 \leq \theta \leq \pi.$

Due to symmetry, any smallest inscribing circle should be centered at some point $(0,h)$ on the positive $y$-axis. Then we can set up the problem in the following way. Obviously in order for every point in the semi-circle centered at $(\frac{\sqrt{2}}{4}, \frac{\sqrt{2}}{4})$ inside a circle centered at $(0,h)$ we would need to have the radius be $$\begin{align*}R(h) &= \max_{\theta \in [0,\pi]} \sqrt{ \left( \frac{\sqrt{2}}{4} + \frac{1}{2} \cos \left( \theta - \frac{\pi}{4} \right) \right)^2 + \left( \frac{\sqrt{2}}{4} - h + \frac{1}{2} \sin \left( \theta - \frac{\pi}{4}\right) \right)^2} \\ &= \max_{\theta \in [0,\pi]} \sqrt{ \frac{1}{8} + \frac{\sqrt{2}}{4} \cos \left( \theta - \frac{\pi}{4} \right) + \frac{1}{4} \cos^2 \left( \theta - \frac{\pi}{4} \right) + \left( \frac{\sqrt{2}}{4} - h \right)^2 + \left( \frac{\sqrt{2}}{4} - h \right) \sin \left( \theta - \frac{\pi}{4} \right) + \frac{1}{4} \sin^2 \left( \theta - \frac{\pi}{4} \right) } \\ &= \max_{\theta \in [0,\pi]} \sqrt{ \frac{3}{8} + \left( \frac{\sqrt{2}}{4} - h \right)^2 + \frac{\sqrt{2}}{4} \cos \left( \theta - \frac{\pi}{4} \right) + \left( \frac{\sqrt{2}}{4} - h \right) \sin \left( \theta - \frac{\pi}{4} \right) }.\end{align*}$$ Since square roots are monotonic and several of the terms are constant with respect to $\theta$, the optimal value of \theta in to obtain $R(h)$ will be equivalent to the optimizer of the function $$f_h(\theta) = \frac{\sqrt{2}}{4} \cos \left( \theta - \frac{\pi}{4} \right) + \left( \frac{\sqrt{2}}{4} - h \right) \sin \left( \theta - \frac{\pi}{4} \right)$$ on $\theta \in [0,\pi].$ Taking the derivative with respect to $\theta$ and setting equal to zero we see that the only critical value of $f_h$ in that interval satisfies $$-\frac{\sqrt{2}}{4} \sin \left( \theta^* - \frac{\pi}{4} \right) + \left( \frac{\sqrt{2}}{4} - h \right) \cos \left( \theta^* - \frac{\pi}{4} \right) = 0,$$ or equivalently $$\tan\left(\theta^* - \frac{\pi}{4}\right) = 1 - 2\sqrt{2}h,$$ which implies that $$\begin{align*}\cos \left(\theta^* -\frac{\pi}{4} \right) &= \frac{1}{\sqrt{1 + (1 - 2\sqrt{2}h)^2}} \\ \sin \left( \theta^* - \frac{\pi}{4} \right) &= \frac{1-2\sqrt{2}h}{\sqrt{1 + (1 - 2\sqrt{2}h)^2}},\end{align*}$$ since $\cos ( \tan^{-1} x ) = \frac{1}{\sqrt{1+x^2}}$ and $\sin ( \tan^{-1} x ) = \frac{x}{\sqrt{1+x^2}}.$ Plugging this back into $f_h,$ we get $$\begin{align*}f_h(\theta^*) &= \frac{\sqrt{2}}{4} \frac{1}{\sqrt{1 + (1-2\sqrt{2}h)^2}} + \left(\frac{\sqrt{2}}{4} - h\right) \frac{1 - 2\sqrt{2}h}{\sqrt{1 + (1-2\sqrt{2}h)^2}} \\ &= \frac{ \sqrt{2} }{4} \frac{ 1 + (1-2\sqrt{2}h)^2 }{ \sqrt{ 1+ (1-2\sqrt{2}h)^2}} \\ &= \frac{\sqrt{2}}{4} \sqrt{1 + (1 - 2\sqrt{2}h)^2}.\end{align*}$$ Plugging this value back into $R(h)$ we get $$R(h) = \sqrt{\frac{3}{8} + \frac{1}{8} (1 - 2\sqrt{2}h)^2 + \frac{\sqrt{2}}{4} \sqrt{1 + (1-2\sqrt{2}h)^2}}.$$

Now in order for the entire unit square to be contained within the circle, the lowest point $(0, -\frac{\sqrt{2}}{2})$ must also be less than or equal to $R(h)$ away from the center at $(0,h).$ In particular, if we want to have the inscribed within the circle we should have the $R(h)$ equal the distance between $(0, -\frac{\sqrt{2}}{2})$ and $(0,h),$ that is $h + \frac{\sqrt{2}}{2}.$ This leads to the equation $$h + \frac{\sqrt{2}}{2} = \sqrt{ \frac{3}{8} + \frac{1}{8} (1- 2\sqrt{2}h)^2 + \frac{\sqrt{2}}{4} \sqrt{ 1 + (1-2\sqrt{2}h)^2}}.$$ Squaring both sides yields the equivalent equation $$\left(h + \frac{\sqrt{2}}{2}\right)^2 = \frac{3}{8} + \frac{1}{8} \left( 1 - 2\sqrt{2}h \right)^2 + \frac{\sqrt{2}}{4} \sqrt{ 1 + (1 - 2\sqrt{2}h)^2 }$$ which is again equivalent to $$\frac{3}{2} \sqrt{2} h = \frac{\sqrt{2}}{4} \sqrt{1 + (1-2\sqrt{2}h)^2 }.$$ Squaring both sides again yield $$\frac{9}{2} h^2 = \frac{1}{8} \left( 1 + (1 - 2\sqrt{2}h)^2 \right) = h^2 - \frac{\sqrt{2}}{2} h + \frac{1}{4}$$ or equivalently $$\frac{7}{2} h^2 + \frac{\sqrt{2}}{2} h - \frac{1}{4} = 0.$$ Solving the quadratic in terms of $h$ leaves the only positive solution as $$h^* = \frac{-\frac{\sqrt{2}}{2} + \sqrt{ \frac{1}{2} - 4 \cdot \frac{7}{2} \cdot -\frac{1}{4}}}{2 \frac{7}{2}} = \frac{ -\frac{\sqrt{2}}{2} + \sqrt{ \frac{1}{2} + \frac{7}{2} }}{7} = \frac{ 2 - \frac{\sqrt{2}}{2} }{7} = \frac{4 - \sqrt{2}}{14}.$$ Therefore, the radius of the smallest circle that contains this heart shape is $$R(h^*) = \frac{4-\sqrt{2}}{14} + \frac{\sqrt{2}}{2} = \frac{4 - \sqrt{2} + 7 \sqrt{2}}{14} = \frac{4 + 6 \sqrt{2}}{14} = \frac{2 + 3\sqrt{2}}{7} \approx 0.89180581\dots.$$