A common tangent to `9x^2-16y^2 = 144` and `x^2 + y^2 = 9`, is

Options

A. `y=(3)/(sqrt(7))x+(15)/(sqrt(17))`
B. `y=3 sqrt((2)/(sqrt(17)))x+(25)/(sqrt(7))`
C. `y=2sqrt((3)/(7))x+15sqrt(7)`
D. `y=-3 sqrt((2)/(sqrt(7)))x +(25)/(sqrt(7))`

Answer

To find the common tangent to the hyperbola \(9x^2 - 16y^2 = 144\) and the circle \(x^2 + y^2 = 9\), we can follow these steps:

### Step 1: Rewrite the equations in standard form
The hyperbola can be rewritten as:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]
This shows that \(a^2 = 16\) and \(b^2 = 9\).

The circle is already in standard form:
\[ x^2 + y^2 = 9 \]
This indicates that the radius of the circle is \(r = 3\).

### Step 2: Assume the equation of the tangent line
Let the equation of the common tangent be:
\[ y = mx + c \]
where \(m\) is the slope of the tangent and \(c\) is the y-intercept.

### Step 3: Use the tangent condition for the hyperbola
For the line to be tangent to the hyperbola, we use the condition:
\[ c^2 = a^2 m^2 - b^2 \]
Substituting the values of \(a^2\) and \(b^2\):
\[ c^2 = 16m^2 - 9 \quad \text{(Equation 1)} \]

### Step 4: Use the tangent condition for the circle
For the line to be tangent to the circle, we use the condition:
\[ \frac{c}{\sqrt{m^2 + 1}} = r \]
Substituting \(r = 3\):
\[ c = 3\sqrt{m^2 + 1} \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Now, we substitute \(c\) from Equation 2 into Equation 1:
\[ (3\sqrt{m^2 + 1})^2 = 16m^2 - 9 \]
This simplifies to:
\[ 9(m^2 + 1) = 16m^2 - 9 \]
Expanding and rearranging gives:
\[ 9m^2 + 9 = 16m^2 - 9 \]
\[ 0 = 7m^2 - 18 \]
\[ 7m^2 = 18 \]
\[ m^2 = \frac{18}{7} \]
Thus, we find:
\[ m = \pm \sqrt{\frac{18}{7}} = \pm \frac{3\sqrt{2}}{\sqrt{7}} \]

### Step 6: Substitute \(m\) back to find \(c\)
Now, we substitute \(m^2\) back into Equation 1 to find \(c^2\):
\[ c^2 = 16\left(\frac{18}{7}\right) - 9 \]
Calculating gives:
\[ c^2 = \frac{288}{7} - 9 = \frac{288}{7} - \frac{63}{7} = \frac{225}{7} \]
Thus:
\[ c = \pm \frac{15}{\sqrt{7}} \]

### Step 7: Write the final equations of the common tangents
The common tangents can now be expressed as:
\[ y = \pm \frac{3\sqrt{2}}{\sqrt{7}}x \pm \frac{15}{\sqrt{7}} \]

### Final Answer
The common tangents to the hyperbola and the circle are:
\[ y = \frac{3\sqrt{2}}{\sqrt{7}}x + \frac{15}{\sqrt{7}} \quad \text{and} \quad y = -\frac{3\sqrt{2}}{\sqrt{7}}x - \frac{15}{\sqrt{7}} \]