Let PQ be the focal chord of the parabola `y^(2)=4x`. If the centre of the circle having PQ as its diameter lies on the line `y=(4)/(sqrt5)`, then the radius (in units) is equal to

Answer

To solve the problem step-by-step, we need to find the radius of the circle whose diameter is the focal chord PQ of the parabola \( y^2 = 4x \), given that the center of the circle lies on the line \( y = \frac{4}{\sqrt{5}} \).

### Step 1: Understand the parabola and focal chord
The equation of the parabola is given by \( y^2 = 4x \). The focus of this parabola is at the point \( (1, 0) \). A focal chord is a line segment that passes through the focus and has its endpoints on the parabola.

### Step 2: Parametric representation of the parabola
The points on the parabola can be represented parametrically as:
- \( P(t_1) = (t_1^2, 2t_1) \)
- \( Q(t_2) = (t_2^2, 2t_2) \)

### Step 3: Relationship between \( t_1 \) and \( t_2 \)
For a focal chord, the relationship between the parameters \( t_1 \) and \( t_2 \) is given by:
\[ t_1 \cdot t_2 = -1 \]
This means that if \( t_1 \) is one parameter, then \( t_2 \) can be expressed as \( t_2 = -\frac{1}{t_1} \).

### Step 4: Find the center of the circle
The center of the circle, which is the midpoint of the segment PQ, can be calculated as follows:
- The x-coordinate of the center is:
\[ \frac{t_1^2 + t_2^2}{2} \]
- The y-coordinate of the center is:
\[ \frac{2t_1 + 2t_2}{2} = t_1 + t_2 \]

### Step 5: Substitute the relationship into the center coordinates
Since \( t_2 = -\frac{1}{t_1} \), we can substitute this into the coordinates:
- The x-coordinate becomes:
\[ \frac{t_1^2 + \left(-\frac{1}{t_1}\right)^2}{2} = \frac{t_1^2 + \frac{1}{t_1^2}}{2} \]
- The y-coordinate becomes:
\[ t_1 - \frac{1}{t_1} \]

### Step 6: Set the y-coordinate equal to \( \frac{4}{\sqrt{5}} \)
We set the y-coordinate equal to the given line:
\[ t_1 - \frac{1}{t_1} = \frac{4}{\sqrt{5}} \]
Multiplying through by \( t_1 \) gives:
\[ t_1^2 - 1 = \frac{4}{\sqrt{5}} t_1 \]
Rearranging this leads to:
\[ t_1^2 - \frac{4}{\sqrt{5}} t_1 - 1 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 1 \), \( b = -\frac{4}{\sqrt{5}} \), and \( c = -1 \).
\[ t_1 = \frac{\frac{4}{\sqrt{5}} \pm \sqrt{\left(-\frac{4}{\sqrt{5}}\right)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
Calculating the discriminant:
\[ \left(-\frac{4}{\sqrt{5}}\right)^2 + 4 = \frac{16}{5} + 4 = \frac{16}{5} + \frac{20}{5} = \frac{36}{5} \]
Thus,
\[ t_1 = \frac{\frac{4}{\sqrt{5}} \pm \frac{6}{\sqrt{5}}}{2} = \frac{10/\sqrt{5}}{2} = \frac{5}{\sqrt{5}} = \sqrt{5} \]
or
\[ t_1 = \frac{-2/\sqrt{5}}{2} = -\frac{1}{\sqrt{5}} \]

### Step 8: Calculate the length of PQ
Using \( t_1 = \sqrt{5} \) and \( t_2 = -\frac{1}{\sqrt{5}} \):
- The length of PQ is given by:
\[ PQ = \sqrt{(t_1^2 - t_2^2)^2 + (2t_1 - 2t_2)^2} \]
Calculating \( PQ \):
\[ t_1^2 = 5, \quad t_2^2 = \frac{1}{5} \]
\[ PQ = \sqrt{(5 - \frac{1}{5})^2 + (2\sqrt{5} + \frac{2}{\sqrt{5}})^2} \]
\[ = \sqrt{(4.8)^2 + (2\sqrt{5} + \frac{2}{\sqrt{5}})^2} \]

### Step 9: Find the radius
The radius \( r \) of the circle is half of the length of the diameter \( PQ \):
\[ r = \frac{PQ}{2} \]

### Final Answer
After calculating the above expressions, we find that the radius \( r \) is equal to \( 3.6 \) units.