The equation `2x^(2)+2y^(2)-6x+8y+k=0` represents a point circle if k is equal to ...........

Answer

To determine the value of \( k \) for which the equation \( 2x^2 + 2y^2 - 6x + 8y + k = 0 \) represents a point circle, we can follow these steps:

### Step 1: Understand the definition of a point circle
A point circle is a circle with a radius of 0. This means that the equation of the circle must satisfy the condition that the radius calculated from the general form of the circle's equation is equal to 0.

### Step 2: Rewrite the given equation
The given equation is:
\[ 2x^2 + 2y^2 - 6x + 8y + k = 0 \]
To compare it with the standard form of a circle, we divide the entire equation by 2:
\[ x^2 + y^2 - 3x + 4y + \frac{k}{2} = 0 \]

### Step 3: Identify coefficients
Now we can identify the coefficients in the standard form of the circle's equation:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
From our equation, we have:
- \( g = -\frac{3}{2} \)
- \( f = 2 \)
- \( c = \frac{k}{2} \)

### Step 4: Use the radius formula
The radius \( r \) of the circle is given by the formula:
\[ r = \sqrt{g^2 + f^2 - c} \]
For a point circle, we set \( r = 0 \):
\[ 0 = \sqrt{g^2 + f^2 - c} \]

### Step 5: Substitute the values of \( g \), \( f \), and \( c \)
Substituting the values we found:
\[ 0 = \sqrt{\left(-\frac{3}{2}\right)^2 + (2)^2 - \frac{k}{2}} \]
Calculating \( g^2 \) and \( f^2 \):
\[ \left(-\frac{3}{2}\right)^2 = \frac{9}{4}, \quad (2)^2 = 4 \]
Thus,
\[ 0 = \sqrt{\frac{9}{4} + 4 - \frac{k}{2}} \]
Converting 4 to a fraction:
\[ 4 = \frac{16}{4} \]
So,
\[ 0 = \sqrt{\frac{9}{4} + \frac{16}{4} - \frac{k}{2}} = \sqrt{\frac{25}{4} - \frac{k}{2}} \]

### Step 6: Square both sides
Squaring both sides gives:
\[ 0 = \frac{25}{4} - \frac{k}{2} \]

### Step 7: Solve for \( k \)
Rearranging the equation:
\[ \frac{k}{2} = \frac{25}{4} \]
Multiplying both sides by 2:
\[ k = \frac{25}{2} \]

### Conclusion
Thus, the value of \( k \) for which the equation represents a point circle is:
\[ \boxed{\frac{25}{2}} \]