Circle tangent to two lines and another circle

In summary, a circle can be tangent to two lines and another circle simultaneously, creating specific geometric configurations. The tangents from points on the circle to the lines form right angles, while the distance from the center of the circle to the lines and the other circle's center is crucial for determining the radius and position of the tangential circle. This scenario often involves applying theorems related to tangents and properties of circles to find intersections and tangential points.
  • #1
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TL;DR Summary
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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  • #2
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.
 
  • #3
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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