High School Circle tangent to two lines and another circle

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SUMMARY

The discussion focuses on determining the center coordinates (x1, y1) of a circle that is tangent to two perpendicular lines at x=a and y=b, as well as a circle centered at the origin with radius r, using OpenSCAD. The user identifies that the distance from the origin to the tangent circle's center must equal the sum of the two radii. A closed-form solution is derived, leading to the quadratic equation R=b-a+r ± √2√((r-a)(b+r)), which allows for the calculation of both the radius R and the coordinates (x1, y1).

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TL;DR
Find center of a circle tangent to perpendicular lines x=a and y=b, and also tangent to a circle of radius r at the origin.
I'm trying to solve this for a model I'm making in OpenSCAD.

Given a circle of radius r centered on the origin, and two perpendicular lines at x=a and y=b, where is the center (x1,y1) of a circle that is tangent to both lines and the centered circle?

Here's a picture:
1708674575479.png


I thought it would be easy, like solving for a circle that intersects 3 points, but there's something I'm not getting here. It's been 4 decades since I had to solve problems like this.

I know that the distance between (0,0) and (x1,y1) should be the sum of the two radii. I could solve it iteratively, but it feels like there should be a closed-form solution here.
 
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You have three unknowns, ##x_1, y_1, r_1## and three equations from the three tangent conditions. You will need to solve a quadratic equation.
 
Thanks, I believe I figured it out.

From the diagram, I have

##x_1^2+y_1^2 = (r+R)^2##

where ##R## is the radius of the unknown circle. I also know that ##x_1=a+R## and ##y_1=b-R##. That means

##(a+R)^2 + (b-R)^2=(r+R)^2##

and I have a quadratic expression with one unknown, ##R##. That can be solved by the quadratic equation, and I get this:

##R=b-a+r \pm \sqrt{2}\sqrt{(r-a)(b+r)}##

With that I can solve for ##(x_1,y_1)##.
 

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