A circle is drawn through the point of intersection of the parabola `y=x^(2)-5x+4` and the x-axis such that origin lies outside it. The length of a tangent to the circle from the origin is ________ .

Explanation

To solve the problem step by step, we need to find the length of the tangent from the origin to a circle that passes through the points of intersection of the parabola \( y = x^2 - 5x + 4 \) with the x-axis.

### Step 1: Find the points of intersection of the parabola with the x-axis.
To find the points where the parabola intersects the x-axis, we set \( y = 0 \):

\[ x^2 - 5x + 4 = 0 \]

### Step 2: Factor the quadratic equation.
Factoring the quadratic gives:

\[ (x - 1)(x - 4) = 0 \]

Thus, the solutions are:

\[ x = 1 \quad \text{and} \quad x = 4 \]

### Step 3: Identify the points of intersection.
The points of intersection on the x-axis are:

\[ (1, 0) \quad \text{and} \quad (4, 0) \]

### Step 4: Find the center of the circle.
The center of the circle, \( C \), is the midpoint of the segment joining the points \( (1, 0) \) and \( (4, 0) \):

\[ C = \left( \frac{1 + 4}{2}, 0 \right) = \left( \frac{5}{2}, 0 \right) \]

### Step 5: Calculate the radius of the circle.
The radius \( r \) of the circle can be calculated as the distance from the center \( C \) to either of the intersection points, say \( (1, 0) \):

\[ r = \sqrt{\left( \frac{5}{2} - 1 \right)^2 + (0 - 0)^2} = \sqrt{\left( \frac{3}{2} \right)^2} = \frac{3}{2} \]

### Step 6: Calculate the distance from the origin to the center of the circle.
The distance \( OC \) from the origin \( (0, 0) \) to the center \( C \left( \frac{5}{2}, 0 \right) \) is:

\[ OC = \sqrt{\left( \frac{5}{2} - 0 \right)^2 + (0 - 0)^2} = \frac{5}{2} \]

### Step 7: Use the Pythagorean theorem to find the length of the tangent.
Using the Pythagorean theorem in triangle \( OTC \):

\[ OT^2 + CT^2 = OC^2 \]

Where \( OT \) is the length of the tangent, \( CT \) is the radius, and \( OC \) is the distance from the origin to the center. Rearranging gives:

\[ OT^2 = OC^2 - CT^2 \]

Substituting the known values:

\[ OT^2 = \left( \frac{5}{2} \right)^2 - \left( \frac{3}{2} \right)^2 \]

Calculating:

\[ OT^2 = \frac{25}{4} - \frac{9}{4} = \frac{16}{4} = 4 \]

Taking the square root gives:

\[ OT = \sqrt{4} = 2 \]

### Final Answer:
The length of the tangent from the origin to the circle is:

\[ \boxed{2} \]