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Home/ Questions/Q 7049911
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T03:04:44+00:00 2026-05-28T03:04:44+00:00

I’m trying to learn canvas by implementing a pie chart. I’ve managed to parse

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I’m trying to learn canvas by implementing a pie chart. I’ve managed to parse my data, draw the slices, and calculate the center of each arc, as noted by the black circles. But now I’m trying to draw one of the slices as though it had been “slid out”. Not animate it (yet), just simply draw the slice as though it had been slid out.

I thought the easiest way would be to first calculate the point at which the new corner of the slice should be (free-hand drawn with the red X), translate there, draw my slice, then translate the origin back. I thought I could calculate this easily, since I know the center of the pie chart, and the point of the center of the arc (connected with a free-hand black line on the beige slice). But after asking this question, it seems this will involve solving a system of equations, one of which is second order. That’s easy with a pen and paper, dauntingly hard in JavaScript.

Is there a simpler approach? Should I take a step back and realize that doing this is really the same as doing XYZ?

I know I haven’t provided any code, but I’m just looking for ideas / pseudocode. (jQuery is tagged in the off chance there’s a plugin will somehow help in this endeavor)

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  1. Editorial Team
    Editorial Team
    2026-05-28T03:04:44+00:00Added an answer on May 28, 2026 at 3:04 am

    Getting the x and y of the translation is easy enough.

    // cx and cy are the coordinates of the centre of your pie
    // px and py are the coordinates of the black circle on your diagram
    // off is the amount (range 0-1) by which to offset the arc
    //      adjust off as needed.
    // rx and ry will be the amount to translate by
    
    var dx = px-cx, dy = py-cy,
        angle = Math.atan2(dy,dx),
        dist = Math.sqrt(dx*dx+dy*dy);
    rx = Math.cos(angle)*off*dist;
    ry = Math.sin(angle)*off*dist;
    

    Plug that into the code Simon Sarris gave you and you’re done. I’d suggest an off value of 0.25.

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