In figure
- There is a Circle with center point C and radius 4
- Points A & B lie on the circle
- AB = 6
- DE is Perpendicular Bisector of AB
- C lies on DE
- DE = 43.5
How to create a circle with radius 3 whose center should lie on DE, and should have same tangents as the first circle?
You have the information that radius is $3$. So you just need the position of centre. Let the centre of new circle be $O$.
Then we can find distance OE using similarity of triangles (not shown in figure, right triangles formed using tangents and radiuses)
$$\frac{OE}{3} = \frac{CE}{4} \tag{1}$$
Since $AD=3$ is known and also the radius $AC=4$, we therefore know $DC = \sqrt{AC^2-AD^2} = \sqrt{7}$.
Also we know $CE = DE-DC = 43.5-\sqrt{7}$. Thus we know the distance of new centre $O$ from $E$ as
$$OE=\frac{3}{4}(43.5-\sqrt{7})$$