Key Concepts and Formulas

• Midpoint Formula: The midpoint of a line segment joining points $(x_1, y_1)$ and $(x_2, y_2)$ is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
• Parametric Equation of a Circle: A point on the circle $x^2 + y^2 = r^2$ can be represented as $(r\cos\theta, r\sin\theta)$, where $\theta$ is a parameter.
• Equation of a Circle: The general equation of a circle is $(x-a)^2 + (y-b)^2 = r^2$, where $(a, b)$ is the center and $r$ is the radius.

Step-by-Step Solution

1. Define Points and the Midpoint

Let $A = (3, 2)$ be the given point. Let $B = (x, y)$ be a point on the circle $x^2 + y^2 = 1$. Let $M = (h, k)$ be the midpoint of the line segment $AB$. We want to find the locus of $M$.

• Why this step? Defining the points and the midpoint allows us to set up the problem using the midpoint formula and the equation of the given circle.

2. Express B in Parametric Form

Since $B$ lies on the circle $x^2 + y^2 = 1$, we can represent $B$ as $(\cos\theta, \sin\theta)$.

• Why this step? Using the parametric form for points on the circle simplifies the calculations and reduces the number of variables.

3. Apply the Midpoint Formula

The midpoint $M(h, k)$ of the line segment joining $A(3, 2)$ and $B(\cos\theta, \sin\theta)$ is given by: $h = \frac{3 + \cos\theta}{2}$ $k = \frac{2 + \sin\theta}{2}$

• Why this step? Applying the midpoint formula establishes the relationship between the coordinates of the midpoint $(h, k)$ and the parameter $\theta$ of the point on the circle.

4. Eliminate the Parameter $\theta$

We need to eliminate $\theta$ to find the locus of $M$. Rearrange the equations from Step 3 to express $\cos\theta$ and $\sin\theta$ in terms of $h$ and $k$: $\cos\theta = 2h - 3$ $\sin\theta = 2k - 2$ Using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$, substitute the expressions for $\cos\theta$ and $\sin\theta$: $(2h - 3)^2 + (2k - 2)^2 = 1$

• Why this step? The identity $\cos^2\theta + \sin^2\theta = 1$ is crucial because it allows us to eliminate the parameter $\theta$, resulting in an equation solely in terms of $h$ and $k$, which represents the locus.

5. Simplify the Equation

Expand and simplify the equation: $(4h^2 - 12h + 9) + (4k^2 - 8k + 4) = 1$ $4h^2 - 12h + 4k^2 - 8k + 13 = 1$ $4h^2 - 12h + 4k^2 - 8k + 12 = 0$ Divide by 4: $h^2 - 3h + k^2 - 2k + 3 = 0$ Replace $(h, k)$ with $(x, y)$ to obtain the equation of the locus: $x^2 - 3x + y^2 - 2y + 3 = 0$

• Why this step? Expanding and simplifying reveals the geometric nature of the locus. Dividing by 4 brings it to a simpler form, making it easier to complete the square or identify the center and radius.

6. Rewrite in Standard Circle Form

Complete the square to rewrite the equation in the standard form of a circle $(x-a)^2 + (y-b)^2 = r^2$: $(x^2 - 3x) + (y^2 - 2y) + 3 = 0$ $\left(x^2 - 3x + \frac{9}{4}\right) + \left(y^2 - 2y + 1\right) + 3 - \frac{9}{4} - 1 = 0$ $\left(x - \frac{3}{2}\right)^2 + (y - 1)^2 = \frac{9}{4} + 1 - 3$ $\left(x - \frac{3}{2}\right)^2 + (y - 1)^2 = \frac{9}{4} + \frac{4}{4} - \frac{12}{4}$ $\left(x - \frac{3}{2}\right)^2 + (y - 1)^2 = \frac{1}{4}$

• Why this step? Rewriting the equation in standard form allows us to easily identify the center and radius of the circle.

7. Determine the Radius

The equation of the locus is $\left(x - \frac{3}{2}\right)^2 + (y - 1)^2 = \frac{1}{4}$. Comparing this to the standard form of a circle, the radius $r$ is given by $r^2 = \frac{1}{4}$, so $r = \sqrt{\frac{1}{4}} = \frac{1}{2}$.

• Why this step? This is the final step to answer the question, which specifically asks for the radius of the locus.

Common Mistakes & Tips

• Algebraic Manipulation: Be meticulous with algebraic manipulations, especially when expanding squares and combining terms.
• Parametric Representation: Remember to use the parametric representation of the circle correctly.
• Completing the Square: Practice completing the square to rewrite the equation of the locus in the standard form of a circle.

Summary

We found the locus of the midpoint of the line segment by expressing the coordinates of a point on the circle parametrically, applying the midpoint formula, eliminating the parameter, and simplifying the resulting equation. The locus is a circle with radius $\frac{1}{2}$.

The final answer is $\boxed{\frac{1}{2}}$, which corresponds to option (B).