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Showing content from https://developer.mozilla.org/en-US/docs/Web/API/Window/screenX below:

Window: screenX property - Web APIs

Baseline Widely available *

The Window.screenX read-only property returns the horizontal distance, in CSS pixels, of the left border of the user's browser viewport to the left side of the screen.

Note: An alias of screenX was implemented across modern browsers in more recent times — Window.screenLeft. This was originally supported only in IE but was introduced everywhere due to popularity.

A number equal to the number of CSS pixels from the left edge of the browser viewport to the left edge of the screen.

In our screenleft-screentop (source code) example, you'll see a canvas onto which has been drawn a circle. In this example we are using Window.screenLeft/Window.screenTop plus Window.requestAnimationFrame() to constantly redraw the circle in the same physical position on the screen, even if the window position is moved.

initialLeft = window.screenLeft + canvasElem.offsetLeft;
initialTop = window.screenTop + canvasElem.offsetTop;

function positionElem() {
  let newLeft = window.screenLeft + canvasElem.offsetLeft;
  let newTop = window.screenTop + canvasElem.offsetTop;

  let leftUpdate = initialLeft - newLeft;
  let topUpdate = initialTop - newTop;

  ctx.fillStyle = "rgb(0 0 0)";
  ctx.fillRect(0, 0, width, height);
  ctx.fillStyle = "rgb(0 0 255)";
  ctx.beginPath();
  ctx.arc(
    leftUpdate + width / 2,
    topUpdate + height / 2 + 35,
    50,
    degToRad(0),
    degToRad(360),
    false,
  );
  ctx.fill();

  pElem.textContent = `Window.screenLeft: ${window.screenLeft}, Window.screenTop: ${window.screenTop}`;

  window.requestAnimationFrame(positionElem);
}

window.requestAnimationFrame(positionElem);

These work in exactly the same way as screenX/screenY.

Also in the code we include a snippet that detects whether screenLeft is supported, and if not, polyfills in screenLeft/screenTop using screenX/screenY.

if (!window.screenLeft) {
  window.screenLeft = window.screenX;
  window.screenTop = window.screenY;
}

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