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I have tough question, which I need for a program I am working on.

I have one circle where I know its center position $(x,y)$ and radius, and one straight line with the formula $y=x+b$ where $b$ is known.

Now I need to find out the center position of another circle so that it touches both the line and the first circle. For this second circle I know only the radius.

Known
Circle 1: Radius $r_1,C_1(x_1,y_1)$
Circle 2: Radius $r_2$
Line: $ly=lx+b$ ($b$ is a known lenght)

What I need to find is $C_2(x_2,y_2)$

At a later point I will also need to find the centerpoint of a circle touching $2$ known circles, but I hope the answer to the first part will help in this as well.

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  • $\begingroup$ The solution is not unique. In fact, once $r_2$ gets large enough, there will be four such circles—the first circle will be internally tangent to two of them. $\endgroup$ – amd Oct 14 '16 at 18:52
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I will provide a hint. The hint is as follows:

Suppose the center of Circle 1 is $O_1$ and center of Circle 2 is $O_2$. Consider the distances from $O_1$ to $l$ and from $O_1$ to $O_2$.

Hope the hint helped you because the rest you do is plug the numbers into the formulae of distances.

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  • $\begingroup$ Its incredible helpfull in that it has changed my way of thinking about the problem, but i am still stuck. Im not sure how i can use the formulae of distance here. As i see it i have a point $O_1$, the distance to $O_2$ ($r_1+r_2$), the distance between $O_2$ and $l$ ($r2$), and the formula for $l$. $\endgroup$ – Black Draco Oct 14 '16 at 14:18
  • $\begingroup$ I found out how to do it. I will update my question when i get back from vacation, to add the final solution :) Thank you :D $\endgroup$ – Black Draco Oct 14 '16 at 15:57
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You can't find a circle of a given radius. What you can do is draw a line from the center of the given circle perpendicular to the given line. Then bisect the line segment formed by the foot of the perpendicular and the point where the line crosses the given circle.

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  • $\begingroup$ You appear to have misunderstood the problem. As long as $r_2$ is large enough, you can always position the second circle so that it’s tangent to both the line and first circle. $\endgroup$ – amd Oct 14 '16 at 18:14

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