Is there any php function that returns an image handle so that i can use it in img src html tag?

Question 1

There are 2 more obvious ways to approach this problem.

Write the rendered image to a file

This is possible using imagejpeg, by passing in a second parameter of the filename, for example:
```
 $im = imagecreatetruecolor(..);
 // ..
 imagejpeg($im, 'image.jpg');
```
Use URL Rewriting

You can always pass through all .jpgs in a special folder (for example /dynamic/image.jpg through to your script which will render it). If you want, you could even take the filename (image.jpg) as a parameter to your script telling it what to render if the image is dynamic.

Obviously, if your image needs to be different per request, #2 is the better option. However, #1 is faster and recommended if your image is static (ie. your imagettftext always writes the same text over the same image).

Question 2

Your PHP page can return an image:

<?php
$filename = "my_image.png";
$image = fopen($filename, 'rb');

header("Content-Type: image/png");
header("Content-Length: " . filesize($filename));
fpassthru($image);

You can then make it read GET argument and return appropriate image. Then you could use it in another pages like this:

<img src="my_image_return_page.php?image=<?=$image_id?>">

This would ofcourse work for PNG image type, for other types you will have to change Content-Type header to appropriate format. You can find available formats here.

Extended answer:

Say you want a script that needs to read image from database according to image id passed.

<?php
//Get image id
$imageId = $_GET['id'];

//You search for your image in database here, and return the location
//Additionally if image with that id wasn't found, you can return location of some image that says "Ooops, we couldn't find image you were looking for".
//Say you save image's location in $imageLocation variable.

//Get pathinfo of your image file
$pathInfo = pathinfo($imageLocation);

$extension = $pathInfo['extension'];
if ($extension == "jpg") {
    $contentType = "jpeg";
} elseif ($extension == "png") {
    $contentType = "png";
} elseif ($extension == "gif") {
    $contentType = "gif";
} else {
    //Handle error here if format isn't recognized
}

//Set content type
header("Content-Type: image/".$contentType);
//Set content length
header("Content-Length: ". filesize($imageLocation));

//This is shorter way then using fopen, but has the same result
echo file_get_contents($imageLocation);