Exploration of tangent circles.
First, I would like to make a script tool
1. Make two circles which are an outer circle and an inner circle.
2. Construct a line through a point on the large circle and the center of the circle.
3. Measure the radius from our smaller circle and then create an identical circle on the circumference of big circle.
4. Make a segment from center of original small circle to intersection between new small circle and a line. Then we will find midpoint of this segment.
5. Construct a perpendicular bisector line of pinl dashed segment and mark the intersection between this line and blue dashed line.
6. This new point of intersection is the center of our Tangent Circle, which is represented as orange color.
7. We can see the two triangles are congruent because of SAS congruence requirement. One side is common and the other side is equal having same 90 degrees from perpendicular bisector.
So. Big triangle including two red segments is equilateral triangle. Subtract the radius of small circles from red segment then the segment would be a radius of Tangent Circle
Click here to GSP Tool and explore
There is another case of Tangent circle. Given 2 green circles, an orange circle is the tangent circle.
Investigation of Tangent Circles and the loci of the centers of the Tangent Circles.
Tangent circle 1 
Tangent circle 2 

Case1. One small circle lies inside 

Case 2. Given circles intersect one another 

Case 3. 2 given circles disjoint 
In these cases, the locus of the centers of the constructed tangent circles is either the ellipse or the htperbola.
When the locus is an ellipse, the sum of distance from fixed two points is constant. These two points are called foci of the ellipse. These two foci are the centers of the given circles. When we check above GSP, we can notice that the sum of the distances from any point of ellipse to those two points is constant, which is AF+BF.
In case of hyperbola, the difference of whose distance is constant. The foci are also the centers of the given circles. We can check the AFBF is constant where points A and B are the centers of given two circles.