circles tangent to intersecting line and each other

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staroid

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I am having trouble finding a way to construct a circle, (circle A) which is tangent to two intersecting lines, and to two other circles, (circles B1 and B2) which are of equal radii, and each of which are tangent to the other, and to one of the lines.

Another way to express this problem is:
Given: two intersecting lines, with a third line bisecting the angle formed between them. Two circles of equal radius are tangent to each other at the bisector, and are each tangent to one of the two lines.
Construct a circle tangent to the two lines and the two circles.

I have tried this using, arbitrarily, two lines intersecting at 90 degrees, and then roughing in a drawing of the solution, and representing lengths with x's, y's, a's and b's, and trying to solve it algebraically.
I run into quadratic equations that are not reducible to something that can be constructed with compass and straight edge.

Any help?
Staroid
 
I am having trouble finding a way to construct a circle, (circle A) which is tangent to two intersecting lines, and to two other circles, (circles B1 and B2) which are of equal radii, and each of which are tangent to the other, and to one of the lines.

Another way to express this problem is:
Given: two intersecting lines, with a third line bisecting the angle formed between them. Two circles of equal radius are tangent to each other at the bisector, and are each tangent to one of the two lines.
Construct a circle tangent to the two lines and the two circles.
Click to expand...

Circles B1 and B2 each have a radius that extends to their respective points of tangency with the “outer” lines. Extend these radii “inward” until they intersect at the bisector line. This is the center of circle A.
 
No, sorry. It's not that simple.
If you extend the radius lines inward, they do indeed intersect, but not at a location where you want the center of the circle A. While it is true that there are 2 possible circles which are tangent both to the two B circles, and to the two intersecting lines, we are looking for the one nearest the point of intersection. Except in one special case, the one you describe is not even the right radius.
 
I'm not sure what you mean by the "right radius" as you did not specify any restrictions on radius.

I think you are also overlooking a solution that meets your original problem statement: the circle that I gave you the center for encompasses both the other two circles. Its radius is slightly greater than the diameters of the two circles. The points of tangency with the two lines are the same points of tangency that the two smaller circles have with those lines -- and hence are also the points of tangency between the circles.

I thought it was an interesting solution -- though apparently not the one you desire. :)
 
Ooops! . . .I overlooked the possibility of a circle encompassing the first two.
Perhaps the problem should have been stated differently.
Try this:
Given: two straight lines, Pm and Pn, intersecting at point P, and forming any angle, and two circles, B1 and B2, of equal radius, tangent to each other at the bisector of that angle, and each tangent to one of the two lines.
Construct the circle not encompassing B1 and B2, but which is which is tangent to B1 and B2, and also to the two lines, and, of the circles, lies nearest point P .

Discussion: Any circle tangent to two intersecting lines will obviously have its center on the bisector of the angle between the lines. And it will have radii meeting the lines perpendicularly, which are parallel with similar radii of the two circles, B1 and B2.
But, to be tangent to some point on each of B1 and B2, there must be a straight line connecting its center with the centers of each of the other two.
If one could find the length of its radius, an arc having a radius of that length, plus the radius of B, could be struck on the bisector, which would solve the problem.
For special cases, such as the two lines meeting at angles of 60, or 120 degrees, the problem is not too difficult. It is also solvable when the angle is 90, and leads to an equation as follows: set x as the radius of B1 or B2, and y as the radius of the mystery circle. Draw some lines connecting obvious points, to form convenient rectangles, triangles and so on, form which to set up equations. Then you will see that there is a right triangle whose hypotenuse is x + y, and for which one side is x, and the other is x + sqrt2(x - y). So, [x + sqrt2(x - y)] squared + x squared = (x + y)squared .
The whole thing resolves out to an equation of the form ax^2 + bxy + cy^2. Setting x or y equal to 1, and applying the quadratic equation gives the solution.

And, I thought it would be easy.
 
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