The locus of the point from which the length of the tangent to the circle `x^(2)+y^(2)-2x-4y+4=0` is 3 units is

Options

A. a)`x^(2)+y^(2)-2x-4y=9=0`
B. b)`x^(2)+y^(2)-2x-4y-4=0`
C. c)`x^(2)+y^(2)-2x-4y-3=0`
D. d)`x^(2)+y^(2)-2x-4y-5=0`

Answer

To find the locus of the point from which the length of the tangent to the circle \(x^2 + y^2 - 2x - 4y + 4 = 0\) is 3 units, we can follow these steps:

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

\[ x^2 + y^2 - 2x - 4y + 4 = 0 \]

We can rearrange this as:

\[ x^2 - 2x + y^2 - 4y + 4 = 0 \]

### Step 2: Complete the Square
Next, we complete the square for the \(x\) and \(y\) terms.

For \(x^2 - 2x\):
\[ x^2 - 2x = (x - 1)^2 - 1 \]

For \(y^2 - 4y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

Substituting these back into the equation gives:

\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 + 4 = 0 \]

This simplifies to:

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

Thus, we can write:

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

### Step 3: Identify the Center and Radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = (1, 2)\)
- Radius \(r = 1\)

### Step 4: Use the Length of the Tangent Formula
The length of the tangent \(L\) from a point \((h, k)\) to a circle with center \((a, b)\) and radius \(r\) is given by:

\[ L = \sqrt{(h - a)^2 + (k - b)^2 - r^2} \]

In our case, we want this length to be \(3\) units. Thus, we set up the equation:

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

### Step 5: Square Both Sides
Squaring both sides, we get:

\[ 9 = (h - 1)^2 + (k - 2)^2 - 1 \]

This simplifies to:

\[ (h - 1)^2 + (k - 2)^2 = 10 \]

### Step 6: Write the Locus Equation
This is the equation of a circle with center \((1, 2)\) and radius \(\sqrt{10}\). The locus of the point from which the length of the tangent to the circle is \(3\) units is given by:

\[ (h - 1)^2 + (k - 2)^2 = 10 \]

### Step 7: Substitute Variables
To express this in terms of \(x\) and \(y\), we substitute \(h\) with \(x\) and \(k\) with \(y\):

\[ (x - 1)^2 + (y - 2)^2 = 10 \]

### Final Step: Rearranging
Expanding this gives:

\[ (x^2 - 2x + 1) + (y^2 - 4y + 4) = 10 \]

Combining like terms leads to:

\[ x^2 + y^2 - 2x - 4y - 5 = 0 \]

### Conclusion
Thus, the locus of the point from which the length of the tangent to the circle is \(3\) units is:

\[ \boxed{x^2 + y^2 - 2x - 4y - 5 = 0} \]