If `x^(2)+y^(2)+6x-4y-12=0` then find the ragne of 2x+y

Explanation

To find the range of \(2x + y\) given the equation \(x^2 + y^2 + 6x - 4y - 12 = 0\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x^2 + y^2 + 6x - 4y - 12 = 0 \]
We can rearrange and complete the square for both \(x\) and \(y\).

### Step 2: Complete the square for \(x\)
For \(x^2 + 6x\):
\[ x^2 + 6x = (x + 3)^2 - 9 \]

### Step 3: Complete the square for \(y\)
For \(y^2 - 4y\):
\[ y^2 - 4y = (y - 2)^2 - 4 \]

### Step 4: Substitute back into the equation
Substituting back into the equation, we have:
\[ (x + 3)^2 - 9 + (y - 2)^2 - 4 - 12 = 0 \]
This simplifies to:
\[ (x + 3)^2 + (y - 2)^2 - 25 = 0 \]
or
\[ (x + 3)^2 + (y - 2)^2 = 25 \]

### Step 5: Identify the center and radius
This represents a circle with center \((-3, 2)\) and radius \(5\).

### Step 6: Parametrize the circle
We can parametrize the circle using:
\[ x = -3 + 5 \cos \theta \]
\[ y = 2 + 5 \sin \theta \]

### Step 7: Find \(2x + y\)
Now we substitute these values into \(2x + y\):
\[ 2x + y = 2(-3 + 5 \cos \theta) + (2 + 5 \sin \theta) \]
This simplifies to:
\[ = -6 + 10 \cos \theta + 2 + 5 \sin \theta \]
\[ = -4 + 10 \cos \theta + 5 \sin \theta \]

### Step 8: Find the range of \(10 \cos \theta + 5 \sin \theta\)
To find the range of \(10 \cos \theta + 5 \sin \theta\), we can express it in the form \(R \cos(\theta - \phi)\), where:
\[ R = \sqrt{10^2 + 5^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \]
The maximum value of \(10 \cos \theta + 5 \sin \theta\) is \(5\sqrt{5}\) and the minimum value is \(-5\sqrt{5}\).

### Step 9: Calculate the range of \(2x + y\)
Thus, the range of \(2x + y\) becomes:
\[ -4 + (-5\sqrt{5}) \quad \text{to} \quad -4 + (5\sqrt{5}) \]
This gives us:
\[ [-4 - 5\sqrt{5}, -4 + 5\sqrt{5}] \]

### Final Answer
The range of \(2x + y\) is:
\[ [-4 - 5\sqrt{5}, -4 + 5\sqrt{5}] \]