What is the radius of the circle
`4x^(2) +4y^(2) -20x +12y -15=0`?

Options

A. 14 units
B. 10.5 units
C. 7 units
D. 3.5 units

Answer

To find the radius of the circle given by the equation \( 4x^2 + 4y^2 - 20x + 12y - 15 = 0 \), we will rewrite the equation in the standard form of a circle, which is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius.

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ 4x^2 + 4y^2 - 20x + 12y - 15 = 0 \]

2. **Divide the entire equation by 4 to simplify:**
\[ x^2 + y^2 - 5x + 3y - \frac{15}{4} = 0 \]

3. **Rearrange the equation:**
\[ x^2 - 5x + y^2 + 3y = \frac{15}{4} \]

4. **Complete the square for the \( x \) terms:**
- Take the coefficient of \( x \) (which is -5), halve it to get \(-\frac{5}{2}\), and then square it:
\[ \left(-\frac{5}{2}\right)^2 = \frac{25}{4} \]
- Add and subtract this value:
\[ x^2 - 5x + \frac{25}{4} - \frac{25}{4} \]

5. **Complete the square for the \( y \) terms:**
- Take the coefficient of \( y \) (which is 3), halve it to get \(\frac{3}{2}\), and then square it:
\[ \left(\frac{3}{2}\right)^2 = \frac{9}{4} \]
- Add and subtract this value:
\[ y^2 + 3y + \frac{9}{4} - \frac{9}{4} \]

6. **Rewrite the equation with completed squares:**
\[ \left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \left(y + \frac{3}{2}\right)^2 - \frac{9}{4} = \frac{15}{4} \]

7. **Combine the constants on the right side:**
\[ \left(x - \frac{5}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 = \frac{15}{4} + \frac{25}{4} + \frac{9}{4} \]
\[ = \frac{49}{4} \]

8. **Identify the radius:**
The equation is now in the form:
\[ \left(x - \frac{5}{2}\right)^2 + \left(y + \frac{3}{2}\right)^2 = \left(\frac{7}{2}\right)^2 \]
Thus, the radius \( r \) is:
\[ r = \frac{7}{2} = 3.5 \]

### Final Answer:
The radius of the circle is \( \frac{7}{2} \) or \( 3.5 \) units.