Three Circles With No Common Points

Number of solutions: 8

SOLUTION METHOD

Circle inversion

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Steps

0. We have three differently sized circles c, d and e that do not intersect or touch each other.
1. By dilating, we change the problem to a point, circle, circle problem. - How to do this is described in steps 1 and 2.
2. Draw lines through the centers of the two larger circles. At the point of intersection with the given lines is the center of a circle with the same radius as the third smallest circle.
3. Draw circles with the same centre as the given circles, but their radius is smaller by the radius of the smallest circle.
4. Draw a circle centered at the point created by dilating the smallest circle with a random radius.
5. By circular inversion, we map two smaller circles onto this circle.
6. Find the tangents of these two circles.
7. By circular inversion over the same circle, we make the tangents of the circles tangent to the dilated circles.
8. Find the centres of the resulting circles.
9. Evaluate which circles are the true solutions of the problem. Their centers are colored red.
10. Adjust their radius by dilation so that they touch the original given lines before their dilation. Find a line passing through their centres and through the centre of one of the given circles. At the point of intersection of the line and the selected circle is the center of a circle with the radius of the small circle from the specification.
11. The resulting circles k₁ and k₂. The circle k₁ has outer contact with all the given circles, and the circle k₂ has inner contact with all the circles.
12. Repeat the same procedure as in steps 1-9. Steps 7 and 8 are combined into step 17.
13. Resulting circles k₃ and k₄. Circle k₃ has inner contact with circles c and e and outer contact with circle d, and circle k₄ has outer contact with circles c and e and inner contact with circle d.
14. All the resulting circles found so far are k₁, k₂, k₃ and k₄.
15. Repeat the same procedure as in steps 1-9. Steps 7 and 8 are combined into step 27.
16. Resulting circles k₅ and k₆. Circle k₅ has an outer contact with circles d and e and an inner contact with circle c, and circle k₆ has an inner contact with circles d and e and an outer contact with circle c.
17. All the resulting circles found so far are k₁, k₂, k₃, k₄, k₅ and k₆.
18. Repeat the same procedure as in steps 1-9. Steps 7 and 8 are combined into step 37.
19. Resulting circles k₇ and k₈. Circle k₇ has inner contact with circles c and d and outer contact with circle e, and circle k₈ has outer contact with circles c and d and inner contact with circle e.
20. All the resulting circles are k₁, k₂, k₃, k₄, k₅, k₆, k₇ and k₈.

SOLUTION METHOD

Loci of points satisfying certain properties

GeoGebra construction

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Steps

0. Three circles named c, d and e which are not touching nor intersecting.
1. We construct the hyperbolas k and p, whose focuses are the centers of the circles c and d.
2. We construct the hyperbolas q and r, whose focuses are the centers of the circles c and e.
3. We construct the hyperbolas s and t, whose focuses are the centers of the circles d and e.
4. We find the intersections of the hyperbolas. Those points will be the centers of the circles.
5. We find the tangent points of the final circles with the circles c, d and e. We do it by finding the line which passes through the intersection of the hyperbolas and the center of one of the assigned circles.
6. In the intersection of this line and the assigned circles lies the tangent point of the final circle with the assigned line.
7. Construct the circles k₁ - k₈.