If the length of the tangent from `(5,4)` to
the circle `x^(2) + y^(2) + 2ky = 0` is 1 then find k.

Answer

To solve the problem, we need to find the value of \( k \) such that the length of the tangent from the point \( (5, 4) \) to the circle defined by the equation \( x^2 + y^2 + 2ky = 0 \) is equal to 1.

### Step-by-step Solution:

1. **Identify the center of the circle:**
The given circle is in the form \( x^2 + y^2 + 2ky = 0 \). We can rewrite this in standard form to identify the center and radius. The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, the center \( (h, k) \) can be derived from the equation:
- \( g = 0 \) (since there is no \( x \) term)
- \( f = k \) (from \( 2ky \))
- Thus, the center of the circle is \( (0, -k) \).

2. **Calculate the radius of the circle:**
The radius \( r \) can be found using the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
In our case:
- \( g = 0 \)
- \( f = k \)
- \( c = 0 \)
Therefore, the radius is:
\[ r = \sqrt{0^2 + k^2 - 0} = |k| \]

3. **Use the length of the tangent formula:**
The length of the tangent \( L \) from a point \( (x_1, y_1) \) to a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]
Substituting the values:
- \( (x_1, y_1) = (5, 4) \)
- \( (h, k) = (0, -k) \)
- \( r = |k| \)

The length of the tangent becomes:
\[ L = \sqrt{(5 - 0)^2 + (4 - (-k))^2 - k^2} \]
Simplifying this:
\[ L = \sqrt{5^2 + (4 + k)^2 - k^2} \]
\[ = \sqrt{25 + (16 + 8k + k^2) - k^2} \]
\[ = \sqrt{25 + 16 + 8k} = \sqrt{41 + 8k} \]

4. **Set the length of the tangent equal to 1:**
According to the problem, the length of the tangent is 1:
\[ \sqrt{41 + 8k} = 1 \]

5. **Square both sides to eliminate the square root:**
\[ 41 + 8k = 1 \]

6. **Solve for \( k \):**
Rearranging gives:
\[ 8k = 1 - 41 \]
\[ 8k = -40 \]
\[ k = -5 \]

Thus, the value of \( k \) is \( \boxed{-5} \).