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 rearrange the equation into the standard form of a circle. The standard form of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius.

We start with the given equation:
\[ 5x^2 + 5y^2 + 4x - 8y = 16 \]

### Step 2: Divide the entire equation by 5
To simplify the equation, we divide every term by 5:
\[ x^2 + y^2 + \frac{4}{5}x - \frac{8}{5}y = \frac{16}{5} \]

### Step 3: Rearrange the equation
Now, we rearrange the equation:
\[ x^2 + y^2 + \frac{4}{5}x - \frac{8}{5}y - \frac{16}{5} = 0 \]

### Step 4: Complete the square for \(x\) and \(y\)
Next, we complete the square for the \(x\) and \(y\) terms.

For \(x\):
1. Take the coefficient of \(x\) which is \(\frac{4}{5}\), halve it to get \(\frac{2}{5}\), and square it to get \(\left(\frac{2}{5}\right)^2 = \frac{4}{25}\).
2. Add and subtract \(\frac{4}{25}\) in the equation.

For \(y\):
1. Take the coefficient of \(y\) which is \(-\frac{8}{5}\), halve it to get \(-\frac{4}{5}\), and square it to get \(\left(-\frac{4}{5}\right)^2 = \frac{16}{25}\).
2. Add and subtract \(\frac{16}{25}\) in the equation.

The equation now becomes:
\[ \left(x + \frac{2}{5}\right)^2 - \frac{4}{25} + \left(y - \frac{4}{5}\right)^2 - \frac{16}{25} = \frac{16}{5} \]

### Step 5: Simplify the equation
Combine the constants:
\[ \left(x + \frac{2}{5}\right)^2 + \left(y - \frac{4}{5}\right)^2 = \frac{16}{5} + \frac{4}{25} + \frac{16}{25} \]
Convert \(\frac{16}{5}\) to have a common denominator of 25:
\[ \frac{16}{5} = \frac{80}{25} \]
Now, the equation becomes:
\[ \left(x + \frac{2}{5}\right)^2 + \left(y - \frac{4}{5}\right)^2 = \frac{80}{25} + \frac{4}{25} + \frac{16}{25} = \frac{100}{25} = 4 \]

### Step 6: Identify the center and radius
From the standard form \((x - h)^2 + (y - k)^2 = r^2\), we can identify:
- Center \((h, k) = \left(-\frac{2}{5}, \frac{4}{5}\right)\)
- Radius \(r = \sqrt{4} = 2\)

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

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