Find the length of the intercept on the straight line 4x-3y-10=0 by the circle `x^(2)+y^(2)-2x+4y-20=0`.

Answer

To find the length of the intercept on the straight line \(4x - 3y - 10 = 0\) by the circle \(x^2 + y^2 - 2x + 4y - 20 = 0\), 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:
\[ x^2 + y^2 - 2x + 4y - 20 = 0 \]
We can rearrange it as:
\[ x^2 - 2x + y^2 + 4y = 20 \]
Next, we 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:
\[ (x - 1)^2 - 1 + (y + 2)^2 - 4 = 20 \]
This simplifies to:
\[ (x - 1)^2 + (y + 2)^2 = 25 \]
Thus, the center of the circle is \((1, -2)\) and the radius \(r = 5\).

### Step 3: Find the Perpendicular Distance from the Center to the Line
Now, we need to find the perpendicular distance from the center of the circle \((1, -2)\) to the line \(4x - 3y - 10 = 0\). The formula for the distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
Here, \(A = 4\), \(B = -3\), \(C = -10\), and \((x_0, y_0) = (1, -2)\).

Substituting these values in:
\[ d = \frac{|4(1) - 3(-2) - 10|}{\sqrt{4^2 + (-3)^2}} = \frac{|4 + 6 - 10|}{\sqrt{16 + 9}} = \frac{|0|}{5} = 0 \]

### Step 4: Interpret the Result
Since the distance \(d = 0\), this means that the center of the circle lies on the line. Therefore, the line is a diameter of the circle.

### Step 5: Calculate the Length of the Intercept (Diameter)
The length of the intercept on the line, which is the diameter of the circle, is given by:
\[ \text{Length of the intercept} = 2 \times r = 2 \times 5 = 10 \]

### Final Answer
Thus, the length of the intercept on the straight line by the circle is:
\[ \boxed{10} \]