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How to calculate the center of a rotated element, after resizing it

I've already asked this same question months ago, but no one was able to answer me, even after making a fully functional example on Plunker, then, I am going to ask it again, and yes, I still have the same problem.
My problem: find the centre of an element who have some rotation in it, after resizing it, to use it as the new pivot of rotation.
In my practical example, it is possible to see the problem in action; I have created two circles to show the problem better. After rotating and resizing the element, it's possible to see how the red and blue circles are apart from each other.
Blue Circle: the "correct" position of the centre, achieved by setting the cx/cy coordinates as the calculated element centre, plus, applying the transform rotate in it. The transform translates the circle to the correct position.
Red Circle: same as the blue circle, minus the transform rotate, these values are the ones used as the rotation pivot for the transform rotate().
My assumptions until here: By applying the transform rotate() in the blue circle, I'm considering the rotation angle in the calculated centre, so all I have to do is replicate the matrix calculations made by the rotate() function. I'm already doing this with the four handles that the user can click to make a rotation, what could go wrong?
My goal: Resize an element with rotation keeping the pivot of rotation in the centre.
I think this answer gave me some info, the math here helped me with the rotation handles starting position, but still, I can't find the right way to calculate the new centre after the resize.
The example was made using D3js + AngularJS v1. I work actively with both, but I am new to the geometry math world.
Again, this is the project on Plunker.

To get the centre of the transformed and rotated element, the most accurate way would probably be to get the browser to calculate it for you.
First create an SVGPoint object to hold our original centre point.
var centre = svg.createSVGPoint();
Initialize this point with the centre of the original object. You can get that by calling getBBox() on the element, and performing a smiple calculation.
var bbox = obj.getBBox();
centre.x = bbox.x + bbox.width / 2;
centre.y = bbox.y + bbox.height / 2;
Next, get the transform matrix from the transform attribute of the transformed object
var matrix = transformedObj.transform.baseVal.consolidate().matrix
Now we can transform our SVGPoint object with this matrix.
var transformedCentre = centre.matrixTransform(matrix);
After this, the x and y properties of transformedCentre should be your transformed centre point.
This should work, but I haven't tested it.

Velocity.js - rotation around center

I'm using Velocity.js for animation.
What's the proper way to move an SVG image, while rotating it around the center at the same time?

You do not need to do per frame animation if you can define the motion using combination of rotations and linear movements.
As it comes out, your issue is that you are not able to fully manage the origin of rotation.
NOTE : The following is applicable to not only velocity but all css transformations in general.
The origin of rotation is the top-left corner of the item to be rotated
Now, if you want to move the object without moving the origin, you can use the translateX, translateY properties. eg
.velocity({ translateX: "+=200", translateY: "25%"})
In order to move the object as well as the origin, you need to set or move its x and y position parameters. eg
.velocity({ x: "+=200", y: "25%" })
Rotation around center
As an example, if you want to rotate an object on its center,you need to
Translate the object by -w/2 and -h/2 where the width and height is w and h.
Rotate the object
Rotation around an external point
In case where you need to rotate the object around an item, simply first translate the object away from its origin by a suitable distance and then rotate it.
This PEN shows both examples where the green rectangle is rotated around its center and the blue one around an external point by combining translation and position correctly.

Velocity.js uses CSS transforms to do it's rotation. Because of this, all you actually need to do is set transform-origin: center; in your CSS. This changes the point around which all actions will happen.
Trying to match the rotational movement and shifting your X and Y accordingly will be a slower animation and more prone to bugs.
More information about transform-origin can be found here.

Understanding rotation and calculating the top left point in KineticJS

I am working on a page where I can view images. I want to create a rotation tool. I've done that, but, it's not working consistently. When I set up the centre point to rotate by, the image jumps slightly, and it gets worse each time. I was experimenting, and, I have code to add a wedge to the top left corner of my top level group ( so, at 0,0 ). If I rotate the image by 45 degrees and drag it so that half of it is off the left edge of my canvas, then I call getAbsolutePosition on the wedge and on the group, I get these values:
layer.getAbsolutePosition()
Object {x: 104.66479545850302, y: 279.2748571151325}
wedge.getAbsolutePosition()
Object {x: 180.2684127179338, y: -73.48773356791764}
I think this means my y position is actually the bottom of the image, which is off screen.
What I want to do, is calculate the absolute position of the middle of my image, when the mouse moves over it, regardless of it's rotation. I have some code that works out points with rotation, which seems like it works at first, almost, but it just gets more and more broken the more I use the tool. I feel like there's something about how Kinetic is tracking these things and what it's reporting, that I am missing. Any hints would be most appreciated. Tutorials I can read are even better ( yes, I've read everything linked from the KineticJS site and searched the web ).
In a nutshell, the question is, if I have an image inside a group, and it's rotated, how do I work out the centre point of the image, taking the rotation in to account, and how do I set the offset so it will rotate from that point, and stay in the same place ?
Thanks

As you've discovered about KinetiJS:
rotation is easy
dragging is easy
dragging+rotation is difficult
After you drag your image you must reset its rotation point (offsetX/offsetY).
KineticJS makes dragging+rotation more difficult than it has to be.
Resetting the offset points of your image will cause KineticJS to automatically move your image (Noooo!!).
That's what's causing your jumping.
The solution to the "jumping" problem:
When you reset the image's rotation point (offsetX/OffsetY) you must also reset the image's X/Y position.
This code resets both XY and Offsets for an image after dragging:
A Demo: http://jsfiddle.net/m1erickson/m9Nw7/
// calc new position and offset
var pos=rect.getPosition();
var size=rect.getSize();
var offset=rect.getOffset();
var newX=pos.x-offset.x+size.width/2;
var newY=pos.y-offset.y+size.height/2;
// reset both position and offset
rect.setPosition([newX,newY]);
rect.setOffset(size.width/2,size.height/2);

Adding parallax to scrolling background

I'm making a canvas game where you travel in a spaceship over an endless repreating background. Right now I'm drawing four instances of the background at different positions based off of the player's x/y position, so they will move with the player.
ctx.translate(ax,ay);
ctx.drawImage(Ibg,Math.round(x/1080)*1080,Math.round(y/720)*720,1080,720);
ctx.drawImage(Ibg,(Math.round(x/1080)*1080)-1080,Math.round(y/720)*720,1080,720);
ctx.drawImage(Ibg,Math.round(x/1080)*1080,(Math.round(y/720)*720)-720,1080,720);
ctx.drawImage(Ibg,(Math.round(x/1080)*1080)-1080,(Math.round(y/720)*720)-720,1080,720);
Translating to ax and ay basically allows objects to scroll with the cameras the player moves, since ax and ay are relative to the player's position. I can create a parallax effect by doing this instead:
ctx.translate(ax*.5,ay*.5);
This makes the background scroll slower than other game objects, like I'd like it to. But I still haven't figured out how to adjust the rest of the code to compensate. As the player moves farther from (0,0) he sees less and less of the background, because it seems to go beyond him at a faster rate. How can I fix this?

As an option to markE's answer you don't need to use a second canvas at all (which is a good option to this).
You can simply use CSS for background image and adjust background position with the amount you need.
Demo here
The essential part is simply these lines:
Background X position where -1 can be replaced with the value you want to move it at.
bgx -= 1;
Then for each loop the background position is updated (Y position is fixed in this example):
canvas.style.backgroundPosition = bgx + 'px -30px'; // set X and Y position
When bgx somehow equals the max width of the image you just reset it to the beginning.

Use 2 canvases -- one placed directly on top of the other
A "background" canvas is on the bottom and animates more slowly.
A "game objects" canvas is on the top and animates more quickly.
That way you can create a parallax effect using different animation speeds for each canvas.

Calculate new width when skewing in canvas

I'm using canvas for a project and I have a number of elements that I'm skewing. I'm only skewing on the y value and just want to know what the new width of the image is after skewing (so I can align it with another canvas element). Check out the code below to see what I mean
ctx.save();
//skew the context
ctx.transform(1,0,1.3,0,0,0);
//draw two images with different heights/widths
ctx.drawImage(image,0,0,42,60);
ctx.drawImage(image,0,0,32,25);
The goal would be to know that the 42 by 60 image was now a X by 60 image so I could do some translating before drawing it at 0,0. It's easy enough to measure each image individually, but I have different skew values and heights/widths throughout the project that need to be align. Currently I use this code (works decently for images between 25 and 42 widths):
var skewModifier = imageWidth*(8/6)+(19/3);
var skewAmount = 1.3; //this is dynamic in my app
var width = (skewModifier*skewAmount)+imageWidth;
As images get wider though this formula quickly falls apart (I think it's a sloping formula not a straight value like this one). Any ideas on what canvas does for skews?

You should be able to derive it mathematically. I believe:
Math.atan(skewAmount) is the angle, in radians, that something is skewed with respect to the origin.
So 1.3 would skew the object by 0.915 radians or 52 degrees.
So here's a red unskewed object next to the same object skewed (painted green). So you have a right triangle:
We know the origin angle (0.915 rads) and we know the adjacent side length, which is 60 and 25 for your two images. (red's height).
The hypotenuse is the long side thats being skewed.
And the opposite side is the triangle bottom - how much its been skewed!
Tangent gets us opposite / adjacent if I recall, so for the first one:
tan(0.915) = opposite / 60, solving for the opposite in JavaScript code we have:
opposite = Math.tan(0.915)*60
So the bottom side of the skewed object starts about 77 pixels away from the origin. Lets check our work in the canvas:
http://jsfiddle.net/LBzUt/
Looks good to me!
The triangle in question of course is the canvas origin, that black dot I painted, and the bottom-left of the red rectangle, which is the original position that we're searching for before skewing.
That was a bit of a haphazard explanation. Any questions?

Taking Simon's fiddle example one step further, so you can simply enter the degrees:
Here's the fiddle
http://jsfiddle.net/LBzUt/33/