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For simplicity, let's not do any shifting and only consider non-negative frequencies.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal and in the following denoted by a single variable $f_s$. The frequency-magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $\omega_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < \omega_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2\omega_c^2}u^2 + \frac{f_s^2}{H^2\omega_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{\omega_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{\omega_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.

For simplicity, let's not do any shifting and only consider non-negative frequencies.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal and in the following denoted by a single variable $f_s$. The frequency-magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $f_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < f_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2f_c^2}u^2 + \frac{f_s^2}{H^2f_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{f_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{f_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.

Let's only consider non-negative frequencies for simplicity.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal and in the following denoted by a single variable $f_s$. The frequency-magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $\omega_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < \omega_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2\omega_c^2}u^2 + \frac{f_s^2}{H^2\omega_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{\omega_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{\omega_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.

For simplicity, let's not do any shifting and only consider non-negative frequencies.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal and in the following denoted by a single variable $f_s$. The frequency-magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $\omega_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < \omega_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2\omega_c^2}u^2 + \frac{f_s^2}{H^2\omega_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{\omega_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{\omega_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.

Let's only consider non-negative frequencies for simplicity.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal, in the following denoted by a single variable $f_s$. The magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $\omega_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < \omega_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2\omega_c^2}u^2 + \frac{f_s^2}{H^2\omega_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{\omega_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{\omega_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.

Let's only consider non-negative frequencies for simplicity.

Let's assume that the horizontal and vertical image dimensions are even integers $W$ and $H$. Looking at the output of a $H\times W$ 2-d DFT of the image, the $u$th column with $u\le W/2$ represents a horizontal frequency of $u/W$ times the horizontal sampling frequency and the $v$th row with $v\le H/2$ represents a vertical frequency of $v/H$ times the vertical sampling frequency. For a square image grid the horizontal and vertical sampling frequencies are equal and in the following denoted by a single variable $f_s$. The frequency-magnitude of a 2-d frequency at bin $u, v$ will then be $\sqrt{(u/W)^2 + (v/H)^2}f_s$.

For a cut-off frequency $\omega_c$ your mask would select frequencies:

$$\sqrt{\left(\frac{u}{W}\right)^2 + \left(\frac{v}{H}\right)^2}f_s < \omega_c\tag{1}$$ $$\Rightarrow\frac{f_s^2}{W^2\omega_c^2}u^2 + \frac{f_s^2}{H^2\omega_c^2}v^2 < 1.\tag{2}$$

That indeed defines an ellipse in coordinates $u, v$.

However, if you consider the actual frequencies $\frac{u}{W}f_s$, $\frac{v}{H}f_s$ as coordinates, then what you have is a circle:

$$\text{Eq. 1}$$ $$\Rightarrow\frac{1}{\omega_c^2}\left(\frac{u}{W}f_s\right)^2 + \frac{1}{\omega_c^2}\left(\frac{v}{H}f_s\right)^2 < 1.\tag{3}$$

To summarize, it depends on how you express your frequencies.