If the locus of the mid-point of the line segment from the point $$(3, 2)$$ to a point on the circle, $$x^2 + y^2 = 1$$ is a circle of radius $$r$$, then $$r$$ is equal to

Let a point on the circle $$x^2 + y^2 = 1$$ be $$(\cos\theta, \sin\theta)$$. The midpoint of the segment from $$(3, 2)$$ to this point is $$\left(\dfrac{3 + \cos\theta}{2},\; \dfrac{2 + \sin\theta}{2}\right)$$.

Let $$h = \dfrac{3 + \cos\theta}{2}$$ and $$k = \dfrac{2 + \sin\theta}{2}$$. Then $$\cos\theta = 2h - 3$$ and $$\sin\theta = 2k - 2$$.

Using $$\cos^2\theta + \sin^2\theta = 1$$, we get $$(2h - 3)^2 + (2k - 2)^2 = 1$$, which simplifies to $$\left(h - \dfrac{3}{2}\right)^2 + (k - 1)^2 = \dfrac{1}{4}$$.

This is a circle with centre $$\left(\dfrac{3}{2}, 1\right)$$ and radius $$r = \dfrac{1}{2}$$.