The locus of mid-points of the chords of the circle `x^2 - 2x + y^2 - 2y + 1 = 0` which are of unit length is :

Options

A. `(x-1)^(2)+(y-1)^(2)=(3)/(4)`
B. `(x-1)^(2)+(y-1)^(2)=2 `
C. `(x-1)^(2)+(y-1)^(2)=(1)/(4)`
D. `(x-1)^(2)+(y-1)^(2)=(2)/(3)`

Answer

To find the locus of mid-points of the chords of the circle \(x^2 - 2x + y^2 - 2y + 1 = 0\) that are of unit length, we will follow these steps:

### Step 1: Rewrite the Circle Equation
First, we need to rewrite the given equation of the circle in standard form. The equation is:

\[ x^2 - 2x + y^2 - 2y + 1 = 0 \]

We can complete the square for \(x\) and \(y\):

\[ (x^2 - 2x) + (y^2 - 2y) + 1 = 0 \]

Completing the square:

\[ (x - 1)^2 - 1 + (y - 1)^2 - 1 + 1 = 0 \]

This simplifies to:

\[ (x - 1)^2 + (y - 1)^2 = 1 \]

### Step 2: Identify the Center and Radius
From the equation \((x - 1)^2 + (y - 1)^2 = 1\), we can identify the center and radius of the circle:

- Center: \((1, 1)\)
- Radius: \(1\)

### Step 3: Consider the Chord of Unit Length
Let \(D(h, k)\) be the midpoint of a chord \(AB\) of the circle. The length of the chord \(AB\) is given as \(1\). The distance from the center \(C(1, 1)\) to the midpoint \(D(h, k)\) can be determined using the Pythagorean theorem.

### Step 4: Apply the Pythagorean Theorem
Using the Pythagorean theorem in triangle \(ACD\):

\[ AC^2 = AD^2 + CD^2 \]

Where:
- \(AC\) is the radius of the circle, which is \(1\).
- \(AD\) is half the length of the chord, which is \(\frac{1}{2}\).
- \(CD\) is the distance from the center to the midpoint \(D\).

Substituting the values:

\[ 1^2 = \left(\frac{1}{2}\right)^2 + CD^2 \]

This simplifies to:

\[ 1 = \frac{1}{4} + CD^2 \]

Thus,

\[ CD^2 = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 5: Express CD in Terms of Coordinates
The distance \(CD\) can also be expressed as:

\[ CD = \sqrt{(h - 1)^2 + (k - 1)^2} \]

Setting this equal to \(\sqrt{\frac{3}{4}}\):

\[ \sqrt{(h - 1)^2 + (k - 1)^2} = \sqrt{\frac{3}{4}} \]

Squaring both sides gives:

\[ (h - 1)^2 + (k - 1)^2 = \frac{3}{4} \]

### Step 6: Replace \(h\) and \(k\) with \(x\) and \(y\)
Since \(D(h, k)\) is the midpoint of the chord, we can replace \(h\) and \(k\) with \(x\) and \(y\):

\[ (x - 1)^2 + (y - 1)^2 = \frac{3}{4} \]

### Final Result
Thus, the locus of mid-points of the chords of the circle that are of unit length is given by:

\[ (x - 1)^2 + (y - 1)^2 = \frac{3}{4} \]