Find the centre and radius of circle `5x^(2)+5y^(2)+4x-8y=16.`

Options

A. `(2/5,4/5)`
B. `(-2/5,4/5)`
C. 3
D. 2

Answer

To find the center and radius of the circle given by the equation \(5x^2 + 5y^2 + 4x - 8y = 16\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
First, we need to make the coefficients of \(x^2\) and \(y^2\) equal to 1. We can do this by dividing the entire equation by 5.

\[ \frac{5x^2 + 5y^2 + 4x - 8y}{5} = \frac{16}{5} \]

This simplifies to:

\[ x^2 + y^2 + \frac{4}{5}x - \frac{8}{5}y = \frac{16}{5} \]

### Step 2: Move the constant to the other side
Next, we rearrange the equation to set it to zero:

\[ x^2 + y^2 + \frac{4}{5}x - \frac{8}{5}y - \frac{16}{5} = 0 \]

### Step 3: Identify coefficients
Now, we can compare this equation with the standard form of a circle:

\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]

From our equation, we can identify:

- \(2g = \frac{4}{5}\) → \(g = \frac{2}{5}\)
- \(2f = -\frac{8}{5}\) → \(f = -\frac{4}{5}\)
- \(c = -\frac{16}{5}\)

### Step 4: Find the center
The center of the circle \((h, k)\) can be found using the formulas:

\[ h = -g = -\left(\frac{2}{5}\right) = -\frac{2}{5} \]
\[ k = -f = -\left(-\frac{4}{5}\right) = \frac{4}{5} \]

Thus, the center of the circle is:

\[ \text{Center} = \left(-\frac{2}{5}, \frac{4}{5}\right) \]

### Step 5: Find the radius
The radius \(r\) can be calculated using the formula:

\[ r = \sqrt{g^2 + f^2 - c} \]

Substituting the values we found:

\[ g^2 = \left(\frac{2}{5}\right)^2 = \frac{4}{25} \]
\[ f^2 = \left(-\frac{4}{5}\right)^2 = \frac{16}{25} \]
\[ c = -\frac{16}{5} = -\frac{80}{25} \]

Now substituting these into the radius formula:

\[ r = \sqrt{\frac{4}{25} + \frac{16}{25} + \frac{80}{25}} = \sqrt{\frac{100}{25}} = \sqrt{4} = 2 \]

### Final Answer
Thus, the center of the circle is \(\left(-\frac{2}{5}, \frac{4}{5}\right)\) and the radius is \(2\).

---