Locus of the mid-point of the line joining (3,2) and point on `(x^2+y^2=1)` is a circle of radius r . Find r

Answer

To find the radius \( r \) of the locus of the mid-point of the line joining the point \( (3, 2) \) and any point on the circle defined by the equation \( x^2 + y^2 = 1 \), we can follow these steps:

### Step 1: Define the Points
Let \( A = (3, 2) \) be the fixed point and \( P = (x, y) \) be any point on the circle \( x^2 + y^2 = 1 \).

### Step 2: Midpoint Formula
The midpoint \( M \) of the line segment joining points \( A \) and \( P \) is given by:
\[ M = \left( \frac{x + 3}{2}, \frac{y + 2}{2} \right) \]
Let \( M = (h, k) \). Therefore, we have:
\[ h = \frac{x + 3}{2} \quad \text{and} \quad k = \frac{y + 2}{2} \]

### Step 3: Express \( x \) and \( y \) in terms of \( h \) and \( k \)
From the equations for \( h \) and \( k \), we can express \( x \) and \( y \) as:
\[ x = 2h - 3 \quad \text{and} \quad y = 2k - 2 \]

### Step 4: Substitute into the Circle Equation
Since \( P \) lies on the circle, it satisfies the equation \( x^2 + y^2 = 1 \). Substituting the expressions for \( x \) and \( y \):
\[ (2h - 3)^2 + (2k - 2)^2 = 1 \]

### Step 5: Expand the Equation
Expanding the left-hand side:
\[ (2h - 3)^2 = 4h^2 - 12h + 9 \]
\[ (2k - 2)^2 = 4k^2 - 8k + 4 \]
Combining these:
\[ 4h^2 - 12h + 9 + 4k^2 - 8k + 4 = 1 \]
\[ 4h^2 + 4k^2 - 12h - 8k + 13 = 1 \]
\[ 4h^2 + 4k^2 - 12h - 8k + 12 = 0 \]

### Step 6: Rearranging the Equation
Dividing the entire equation by 4:
\[ h^2 + k^2 - 3h - 2k + 3 = 0 \]
Rearranging gives:
\[ h^2 - 3h + k^2 - 2k + 3 = 0 \]

### Step 7: Completing the Square
Completing the square for \( h \) and \( k \):
\[ (h^2 - 3h + \frac{9}{4}) + (k^2 - 2k + 1) = 0 - 3 + \frac{9}{4} + 1 \]
\[ \left(h - \frac{3}{2}\right)^2 + (k - 1)^2 = \frac{9}{4} - 3 + 1 = \frac{9}{4} - \frac{12}{4} + \frac{4}{4} = \frac{1}{4} \]

### Step 8: Identify the Circle's Radius
The equation of the locus is:
\[ \left(h - \frac{3}{2}\right)^2 + (k - 1)^2 = \left(\frac{1}{2}\right)^2 \]
This shows that the locus is a circle with center \( \left(\frac{3}{2}, 1\right) \) and radius \( \frac{1}{2} \).

### Final Answer
Thus, the radius \( r \) of the circle is:
\[ r = \frac{1}{2} \]