A common tangent to `x^(2) - 2y^(2) =18` and `x^(2) + y^(2) = 9` is

Options

A. `y=2x+3sqrt(5)`
B. `y=sqrt(2x)+3sqrt(3)`
C. `y=2x+3sqrt(7)`
D. `y=sqrt(2x)+3sqrt(5)`

Answer

To find the common tangent to the hyperbola \(x^2 - 2y^2 = 18\) and the circle \(x^2 + y^2 = 9\), we can follow these steps:

### Step 1: Rewrite the equations in standard form
First, we rewrite the equations of the hyperbola and the circle in standard form.

1. **Hyperbola**:
\[ \frac{x^2}{18} - \frac{y^2}{9} = 1 \]
Here, \(a^2 = 18\) and \(b^2 = 9\).

2. **Circle**:
\[ x^2 + y^2 = 9 \quad \text{(which is the same as } \frac{x^2}{9} + \frac{y^2}{9} = 1\text{)} \]
Here, \(r^2 = 9\) (so \(r = 3\)).

### Step 2: Find the equation of the tangent to the hyperbola
The equation of the tangent to the hyperbola \(x^2/a^2 - y^2/b^2 = 1\) is given by:
\[ y = mx \pm \sqrt{a^2m^2 - b^2} \]
Substituting \(a^2 = 18\) and \(b^2 = 9\), we get:
\[ y = mx \pm \sqrt{18m^2 - 9} \]

### Step 3: Find the condition for the tangent to the circle
For the circle \(x^2 + y^2 = r^2\), the equation of the tangent line \(y = mx + c\) must satisfy:
\[ c^2 = r^2(1 + m^2) \]
Substituting \(r^2 = 9\), we have:
\[ c^2 = 9(1 + m^2) \]

### Step 4: Equate the two expressions for \(c\)
From the hyperbola, we have:
\[ c = \pm \sqrt{18m^2 - 9} \]
From the circle, we have:
\[ c = \pm 3\sqrt{1 + m^2} \]
Setting these equal gives:
\[ \sqrt{18m^2 - 9} = 3\sqrt{1 + m^2} \]

### Step 5: Square both sides to eliminate the square root
Squaring both sides results in:
\[ 18m^2 - 9 = 9(1 + m^2) \]
Expanding and simplifying:
\[ 18m^2 - 9 = 9 + 9m^2 \]
\[ 18m^2 - 9m^2 = 9 + 9 \]
\[ 9m^2 = 18 \]
\[ m^2 = 2 \quad \Rightarrow \quad m = \pm \sqrt{2} \]

### Step 6: Substitute \(m\) back to find \(c\)
Now, substituting \(m = \sqrt{2}\) into the equation for \(c\):
\[ c = \sqrt{18(2) - 9} = \sqrt{36} = 6 \]
Thus, we have two cases for \(c\):
1. \(c = 6\)
2. \(c = -6\)

### Step 7: Write the equations of the tangents
The equations of the tangents are:
1. For \(m = \sqrt{2}\):
\[ y = \sqrt{2}x + 6 \]
2. For \(m = \sqrt{2}\):
\[ y = \sqrt{2}x - 6 \]
3. For \(m = -\sqrt{2}\):
\[ y = -\sqrt{2}x + 6 \]
4. For \(m = -\sqrt{2}\):
\[ y = -\sqrt{2}x - 6 \]

### Final Answer
The common tangents to the hyperbola and the circle are:
1. \(y = \sqrt{2}x + 6\)
2. \(y = \sqrt{2}x - 6\)
3. \(y = -\sqrt{2}x + 6\)
4. \(y = -\sqrt{2}x - 6\)