Find the length of intercept on the straight line `2x - y = 5` by the circle `x^(2) + y^(2) - 6x + 8y - 5 = 0 `

Answer

To find the length of the intercept on the straight line \(2x - y = 5\) by the circle given by the equation \(x^2 + y^2 - 6x + 8y - 5 = 0\), we will follow these steps:

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

\[ x^2 + y^2 - 6x + 8y - 5 = 0 \]

We can rearrange this to group the \(x\) and \(y\) terms:

\[ x^2 - 6x + y^2 + 8y = 5 \]

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

For \(x^2 - 6x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

For \(y^2 + 8y\):
\[ y^2 + 8y = (y + 4)^2 - 16 \]

Substituting these back into the equation, we get:

\[ (x - 3)^2 - 9 + (y + 4)^2 - 16 = 5 \]

This simplifies to:

\[ (x - 3)^2 + (y + 4)^2 = 30 \]

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

- Center \(C(3, -4)\)
- Radius \(r = \sqrt{30}\)

### Step 4: Find the Distance from the Center to the Line
We need to find the distance \(P\) from the center of the circle \(C(3, -4)\) to the line \(2x - y - 5 = 0\). The formula for the distance from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is given by:

\[ P = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]

Substituting \(A = 2\), \(B = -1\), \(C = -5\), and the coordinates of the center \(C(3, -4)\):

\[ P = \frac{|2(3) - 1(-4) - 5|}{\sqrt{2^2 + (-1)^2}} = \frac{|6 + 4 - 5|}{\sqrt{4 + 1}} = \frac{|5|}{\sqrt{5}} = \sqrt{5} \]

### Step 5: Calculate the Length of the Intercept
The length of the intercept \(L\) on the line is given by the formula:

\[ L = 2\sqrt{r^2 - P^2} \]

Substituting \(r = \sqrt{30}\) and \(P = \sqrt{5}\):

\[ L = 2\sqrt{30 - 5} = 2\sqrt{25} = 2 \times 5 = 10 \]

### Final Answer
Thus, the length of the intercept on the straight line by the circle is:

\[ \boxed{10} \]