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 Equation
First, we need to rewrite the circle equation in standard form. The given equation is:

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

We can rearrange this to:

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

Now, we complete the square for both \( x \) and \( y \).

### Step 2: Completing 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:

\[ (x - 1)^2 + (y - 2)^2 - 5 = 4 \]

Thus, we have:

\[ (x - 1)^2 + (y - 2)^2 = 9 \]

This shows that the circle has a center at \( (1, 2) \) and a radius of \( 3 \).

### Step 3: Use the Length of Tangent Formula
The formula for the length of the tangent \( L \) from a point \( (x_1, y_1) \) to a circle centered at \( (h, k) \) with radius \( r \) is given by:

\[ L^2 = (x_1 - h)^2 + (y_1 - k)^2 - r^2 \]

Here, \( (x_1, y_1) = (2, -3) \), \( (h, k) = (1, 2) \), and \( r = 3 \).

### Step 4: Substitute the Values
Now, substituting the values into the formula:

\[ L^2 = (2 - 1)^2 + (-3 - 2)^2 - 3^2 \]

Calculating each term:

- \( (2 - 1)^2 = 1^2 = 1 \)
- \( (-3 - 2)^2 = (-5)^2 = 25 \)
- \( 3^2 = 9 \)

Now, substituting these values into the equation:

\[ L^2 = 1 + 25 - 9 \]

### Step 5: Simplify
Now, simplify:

\[ L^2 = 26 - 9 = 17 \]

Thus, the square of the length of the tangent from the point \( (2, -3) \) to the circle is:

\[ \boxed{17} \]

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