B Circle tangent to two lines and another circle

AI Thread Summary
The discussion focuses on finding the center of a circle tangent to two perpendicular lines and another circle in OpenSCAD. The problem involves determining the coordinates (x1, y1) of the tangent circle, given its radius r and the lines at x=a and y=b. The user realizes that the distance from the origin to the center of the tangent circle must equal the sum of the radii. A quadratic equation is derived from the tangent conditions, leading to a solution for the radius R of the tangent circle. Ultimately, the center coordinates can be calculated using the derived expression for R.
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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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