The square of the length of tangent from (2,- 3) on the circle `x^(2)+y^(2)-2x-4y-4=0` is :

Options

A. 15
B. 17
C. 20
D. 30

Explanation

To find the square of the length of the tangent from the point (2, -3) to the circle given by the equation \(x^2 + y^2 - 2x - 4y - 4 = 0\), we can follow these steps:

### Step 1: Rewrite the Circle's Equation
First, we need to rewrite the equation of the circle in standard form. The given equation is:
\[ x^2 + y^2 - 2x - 4y - 4 = 0 \]
We can rearrange this as follows:
\[ x^2 - 2x + y^2 - 4y = 4 \]
Now, we will complete the square for both \(x\) and \(y\).

### Step 2: Complete the Square
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:
\[ (x - 1)^2 - 1 + (y - 2)^2 - 4 = 4 \]
Simplifying this, we have:
\[ (x - 1)^2 + (y - 2)^2 - 5 = 4 \]
Thus, the equation of the circle in standard form is:
\[ (x - 1)^2 + (y - 2)^2 = 9 \]
This indicates that the center of the circle is \((1, 2)\) and the radius is \(3\).

### Step 3: Use the Length of Tangent Formula
The formula for the length of the tangent from a point \((x_1, y_1)\) to a circle \((x - h)^2 + (y - k)^2 = r^2\) is given by:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]
Here, \((x_1, y_1) = (2, -3)\), \((h, k) = (1, 2)\), and \(r = 3\).

### Step 4: Substitute Values into the Formula
Now, substituting the values into the formula:
\[ L = \sqrt{(2 - 1)^2 + (-3 - 2)^2 - 3^2} \]
Calculating each term:
\[ L = \sqrt{(1)^2 + (-5)^2 - 9} \]
\[ L = \sqrt{1 + 25 - 9} \]
\[ L = \sqrt{17} \]

### Step 5: Find the Square of the Length of the Tangent
The square of the length of the tangent is:
\[ L^2 = 17 \]

Thus, the final answer is:
\[ \boxed{17} \]