# Point on circumference a given distance from another point

I am writing a game and need to figure out some math. If I have a circle with the equation $r^2 = (x-d)^2+(y-e)^2$, where $r$, $d$, and $e$ are constants, and a point $A(a,b)$, how do I find the point(s) on the circumference of the circle that are a given straight-line distance $D$ from $A$?

I'm in high school, so I don't have the most extensive math knowledge. However, my first attempt was to find the $x$-values by isolating the $y$-value in the circle equation and substituting it into the distance formula. Then, I would simply substitute the $x$-values back into the circle equation to get the corresponding $y$-values. However, I can't isolate $y$.

In the end, I will only use this where $A$ is on the circumference of the circle already.

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the equation of the circle is with $r^2$ – Wolphram jonny Apr 24 '13 at 0:29
For the sake of clarification, are you looking for the "straight-line distance" from A or the distance along the circumference of the circle from A? – RecklessReckoner Apr 24 '13 at 0:39
Welcome to Math.SE. You've done a nice job explaining the question, its context, and what you've done so far. Now we will work hard to try to solve it. – vadim123 Apr 24 '13 at 0:39
Straight-line distance is what I meant, sorry. – user21760 Apr 24 '13 at 0:43
I asked because the problem for distance along the circumference is easier.... – RecklessReckoner Apr 24 '13 at 0:55

## 1 Answer

If you're looking for the "straight-line distance" $D$* from A to point(s) on the circumference, you would construct a second circle centered on point $A(a,b)$ with a radius of $D$, $D^2 = (x - a)^2 + (y - b)$. You would then set the equations for the circles equal to each other:

$$(x - a)^2 + (y - b) - D^2 = (x - d)^2 + (y - e)^2 - r^2 .$$

*I'm changing the letter to $D$, since you are calling the x-coordinate of the center of A's circle $d$.

This gives you information about the intersection points of the two circles. (One of the nice features of this is that the quadratic terms $x^2$ and $y^2$ will cancel out.) You will now have a relation between $x$ and $y$ that you can substitute into either circle equation to solve for one of the coordinates; the relation will then let you find the other. (There will be two, one, or no solutions, according to the number of intersection points.)

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You are right, I didn't realize, subtracting the equations is much more simple than adding them! I deleted my answer – Wolphram jonny Apr 24 '13 at 0:49
And I had to run back and edit because either way we choose to write this one equation won't produce enough information alone. – RecklessReckoner Apr 24 '13 at 0:51
This makes sense, thanks! – user21760 Apr 24 '13 at 0:58