The $x$-coordinate of the center of mass should be

$$\frac{1}{\frac{\pi}{4}}\int_0^1 x\sqrt{1-x^2}dx$$

because the area of this disc equals $\frac{\pi}{4}$ and $y=\sqrt{1-x^2}$.

Let $x=\sin{t}$, then the integral part above without the coefficient equals

$$\begin{align*} & \int_0^{\frac{\pi}{2}} \sin{t}\cos{t} (d\sin{t})\\ =& \int_0^\frac{\pi}{2} \sin{t}\cos^2{t} dt \\ =&\int_0^{\frac{\pi}{2}}\left(\sin{t} - \sin^3{t}\right)dt\end{align*}$$

The triple angle trigonometry identity gives

$$\sin{3t} = 3\sin{t} - 4\sin^3{t}\implies \sin^3{t}=\frac{1}{4}\left(3\sin{t} - \sin{3t}\right)$$

Continuing the previous integral by setting this result gives

$$\begin{align*} &\int_0^{\frac{\pi}{2}}\left(\sin{t} - \sin^3{t}\right)dt \\ =&\frac{1}{4}\int_0^{\frac{\pi}{2}}\left(4\sin{t} - (3\sin{t} - \sin{3t}) \right)dt \\ =&\frac{1}{4}\int_0^{\frac{\pi}{2}}(\sin t + \sin 3t)dt \\ =&\left.\frac{1}{4}\left(-\cos t - \frac{1}{3}\cos 3t\right)\right| _{0}^{\frac{\pi}{2}}\\=&\frac{1}{3}\end{align*}$$

Therefore, the $x$-coordinate is $\frac{4}{3\pi}$. By symmetry, the $y$-coordinate is the same. Therefore, the final answer is

$$\boxed{\left(\frac{4}{3\pi}, \frac{4}{3\pi}\right)}$$