Find point at which secant crosses circumference of circle

I don't know the proper terminology, or even how to draw the right diagram, but I'm looking to work out a for any given y.

• a = the thick, horizontal line, distance from y to circle's edge
• x = can be considered the x axis
• y = can be considered the y axis
• pink line is the secant (I love Wikipedia)

So a must get shorter, the further down y it is drawn. The second, lower thick horizontal line a shows this.

I know a (I know the radius of the circle) and I know the length of the second, lower line a (because I know the distance along x from the center to the pink line).

I want to find out V, that is, a vector describing where the pink line bisects the circle. My idea is that (I'm a programmer not a mathematician) as line a moves 'south' along y, it must shorten according to some ratio or function of the radius of the circle.

So y and a are related by some function of the radius. Probably.

If someone knows that function, then I can work out y for any given a. So knowing x, I can then work out y in terms of x and arrive at V.

If someone has a better way to work out V, then let me know. In real life, the pink line is at an angle, but I will rotate the entire coordinate space, work out V and then rotate it back :)

Excuse my ignorance.

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You have two $a$'s that do not look to be the same. Please give one (probably the lower one) a different name. – Ross Millikan Mar 26 '13 at 17:09
@RossMillikan Is there anything else wrong? I'd rather correct the diagram once than 5 times and keep uploading a pic and checking the comments, heuristically. – Luke Puplett Mar 26 '13 at 17:15

Let the upper $a$ be $r$, the radius of the circle. The length of the diagonal blue line is then also $r$. The coordinates of $V$ are $(V_x,V_y)$ By the Pythagorean theorem we have $V_x^2+V_y^2=r^2$ and $V_x=a$, so $V=(a,\sqrt{r^2-a^2})$

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I'll change my diagram so upper a is r. – Luke Puplett Mar 26 '13 at 17:30
I changed it back. With an updated diagram, my entire question needs revising! Let's leave it as is. – Luke Puplett Mar 26 '13 at 17:37

Maybe I dign't get it, so sorry!

But if you know $y$ and the radius, say $r$, then from Pythagorean theorem follows that $a=\sqrt{r^2 - y^2}$ and the coordinate of the vector $V$ in the coordinate system ${X,Y}$ in the picture would be $V=(a,y)$ (I'm assuming the orientation of the $Y$-axis is downward (as suggests the arrow).

Sorry if I misunderstood something!

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Sorry i didn't mean to duplicate an answer, when i started editing there was no answers at all, and my browser didn't refresh! – sciamp Mar 26 '13 at 17:25
Sorry if I wasn't clear. I wasn't. I don't know y but I do know both as. – Luke Puplett Mar 26 '13 at 17:28