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Step-by-Step Solution
Step 1: Represent the midpoint of the line segment
Let the fixed point be $A(3, 2)$ and any point on the circle $x^2 + y^2 = 1$ be $B(\cos \theta, \sin \theta)$. The midpoint $P$ of segment $AB$ is given by
$$
P(h, k) = \left(\frac{3 + \cos \theta}{2}, \frac{2 + \sin \theta}{2}\right).
$$
Hence,
$$
h = \frac{3 + \cos \theta}{2}, \quad k = \frac{2 + \sin \theta}{2}.
$$
Step 2: Express cos(θ) and sin(θ) in terms of h and k
Rearranging the above expressions, we get
$$
\cos \theta = 2h - 3, \quad \sin \theta = 2k - 2.
$$
Step 3: Use the Pythagorean identity
Since $\cos^2 \theta + \sin^2 \theta = 1$, substitute the expressions for $\cos \theta$ and $\sin \theta$:
$$
(2h - 3)^2 + (2k - 2)^2 = 1.
$$
Step 4: Expand and simplify
Expand each square:
$$
(2h - 3)^2 = 4h^2 - 12h + 9,
$$
$$
(2k - 2)^2 = 4k^2 - 8k + 4.
$$
So,
$$
4h^2 - 12h + 9 + 4k^2 - 8k + 4 = 1.
$$
Combine like terms:
$$
4h^2 + 4k^2 - 12h - 8k + 13 = 1 \quad \Rightarrow \quad 4h^2 + 4k^2 - 12h - 8k + 12 = 0.
$$
Divide throughout by 4:
$$
h^2 + k^2 - 3h - 2k + 3 = 0.
$$
Step 5: Convert to standard form of a circle
Complete the squares for $h$ and $k$:
$$
h^2 - 3h + k^2 - 2k + 3 = 0.
$$
Rewrite it as:
$$
(h^2 - 3h) + (k^2 - 2k) + 3 = 0.
$$
To complete the square:
$$
h^2 - 3h = \bigl(h^2 - 3h + \frac{9}{4}\bigr) - \frac{9}{4} = (h - \frac{3}{2})^2 - \frac{9}{4},
$$
$$
k^2 - 2k = \bigl(k^2 - 2k + 1\bigr) - 1 = (k - 1)^2 - 1.
$$
So the equation becomes:
$$
(h - \tfrac{3}{2})^2 - \tfrac{9}{4} + (k - 1)^2 - 1 + 3 = 0,
$$
$$
(h - \tfrac{3}{2})^2 + (k - 1)^2 + \left(- \tfrac{9}{4} - 1 + 3\right) = 0.
$$
Simplify the constant terms:
$$
-\tfrac{9}{4} - 1 + 3 = -\tfrac{9}{4} - \tfrac{4}{4} + \tfrac{12}{4} = -\tfrac{13}{4} + \tfrac{12}{4} = -\tfrac{1}{4}.
$$
Hence the equation is:
$$
(h - \tfrac{3}{2})^2 + (k - 1)^2 = \tfrac{1}{4}.
$$
Step 6: Identify the radius
The standard form of a circle is $(h - h_0)^2 + (k - k_0)^2 = r^2$, so the radius $r$ is
$$
r = \frac{1}{2}.
$$
Final Answer
The radius of the locus of the midpoint is $\boxed{\tfrac{1}{2}}$.
