Locus of a circle center?
I put this up a couple of weeks ago, but received no correct answers. Can we try again?
Two fixed, intersecting circles have unequal radii. A variable circle is tangent to both of the fixed circles. Describe the locus of the center of the variable circle.
- gianlinoLv 710 years agoFavourite answer
The difference of the distances to the centers is | R - R'| . So you get a hyperbola.
One branch corresponds to the varying circle containing both fixed,
The other branch to circles either in the intersection or exterior to both.
The asymptotes are normal to the common tangents and cross at the midle of the centers.
Edit: Btw I forgot an ellipse. This one takes into account the circles which are interior to one of the circles and exterior to the other. This ellipse has same foci as the hyperbola so it is orthogonal to it. They cross at 4 points, 2 of them being the intersection of the circles.
@ Madhukar. Obviously you have something inside the intersection. The circles squeezed in there have their centers just there.
You also have the second branch.
This branch is the center of circles containing both intersecting circles.Source(s): http://mathworld.wolfram.com/Hyperbola.html
- MadhukarLv 710 years ago
I have reached the same conclusion as gianlino worked out as under.
Let (a, b) be the center and r1 = radius of the first circle
and (c, d) be the center and r2 = radius of the second circle.
Let (h, k) be the center and r = radius of the variable circle which touches the above two circles.
=> distance of the center of the variable circle from the center of the first given circle = r + r1
and distance of the center of the variable circle from the center of the second given circle = r + r2
=> √[(h - a)^2 + (k - b)^2] = r + r1
and √[(h - c)^2 + (k - d)^2] = r + r2
l √[(h - a)^2 + (k - b)^2] - √[(h - c)^2 + (k - d)^2] l = l r1 - r2 l
Replacing (h, k) by (x, y) gives the locus as
l √[(x - a)^2 + (y - b)^2] - √[(x - c)^2 + (y - d)^2] l = l r1 - r2 l
=> Any point P (x, y) on the locus of the curve is such that the positive difference of its distances from the centers of the given circles = positive difference of the lengths of the radii of the circles.
=> the curve is a hyperbola whose foci are the centers of the given circles and the length of whose transverse axis = (1/2) l r1 - r2 l.
In fact, from the hyperbola, a part is missing lying between the intersecting points of the given two circles. So the set of points on the locus does not form the complete hyperbola, but they lie on the hyperbola given by the above equation.
Also, the points on the locus will lie only on one of the two arms of the hyperbola and not both.
In the following Wolfram Alpha link, two circles with unequal radii and the locus of the centers of the circles tangent to both is shown. The points on one arm of the hyperbola in the link outside the two circles form the locus.
I read your remarks.
I had the circles touching given circles externally in mind. As the asker has not mentioned it, I agree we should consider circles touching the given circles internally as well as touching one internally and the other externally. Considering this, I can visualize that the points on hyperbola between the intersecting points will get covered. However, I could not see how the second branch of hyperbola also may be part of the locus.
- 10 years ago
The variable circle center will form an isosceles triangle with the other two