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    <p>Using the ol' <a href="http://en.wikipedia.org/wiki/Pythagorean_theorem" rel="nofollow noreferrer">Pythagorean Thereom</a>, this simple bit of algebra follows... <img src="https://i.stack.imgur.com/eAioX.png" alt="y in terms of x and r"></p> <p><strong>r</strong> is the radius of the circle.<br> <strong>x</strong> is a variable you'll make iterate through the circle, in increments you can easily work out from knowing the radius and the spacing.</p> <p>Knowing the height of each line (and their x position), it won't be a challenge centering the verticle lines parallel to the cirlce center (if your placement coordinates specify the top left corner as they usually do, you'll want to place it at <code>y = circle_center - (line_length / 2)</code></p> <p>Note also, that if you do specify the coordinates of the vertical lines via the top left corner, you'll need to consider how thick they are when calculating the x coordinate.<br> This is extremely simple, using the x value suggested by the circle.<br> <code>true_x = suggested_x - (line_width / 2)</code></p> <p><strong>Refer to other more sensible answers for dealing with angled lines</strong> </p> <p><strong>This procedure remains as a demonstration of the sensibility of using a clip over finding the exact domain/range of each line</strong> (and because I wasted so much of my life on it).</p> <p><img src="https://i.stack.imgur.com/kPV6Z.png" alt="enter image description here"> <img src="https://i.stack.imgur.com/L6iJ3.png" alt="enter image description here"> <img src="https://i.stack.imgur.com/Py6Pq.png" alt="enter image description here"> <img src="https://i.stack.imgur.com/mgZEE.png" alt="enter image description here"> Here's where the fun ended for me. <img src="https://i.stack.imgur.com/mrsrO.png" alt="enter image description here"></p> <p>Oomph! Replace that last m with 'tan(theta)'</p>
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    1. PODraw circle with lines in it using Java graphics
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    1. COBut what if the line can be angled. The line won't always be vertical, but rather defined by some angle theta.
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    2. COImagine two identical graphs, excepting that one has angled lines. Both graphs will have the same number of lines and the same area between them (mathematically. In pixels, things are yuck). Before, you iterated across the line y = 0 (the x axis) and made perpendicular lines (x = constant). In this case, you can iterate across a line y = mx (where your gradient m - tan(90 - angle)). Your lines will be perpendicular to this quasi-axis (thus at angle). I'll write out a thorough explanation soon (this is a brainstorm for those who can write it out quicker)
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    3. COAdding images as I make them, hopefully they inspire some ideas.
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