Let `y=mx+c` be a common tangent to `(x^(2))/(16)-(y^(2))/(9)=1 and (x^(2))/(4)+(y^(2))/(3)=1`, then find the value of `m^(2)+c^(2)`.

Answer

To find the value of \( m^2 + c^2 \) where \( y = mx + c \) is a common tangent to the hyperbola \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \) and the ellipse \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \), we can follow these steps:

### Step 1: Identify the equations of the curves
The hyperbola is given by:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]
The ellipse is given by:
\[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \]

### Step 2: Write the equation of the tangent line
The equation of the tangent line can be expressed as:
\[ y = mx + c \]

### Step 3: Find the condition for the tangent to the hyperbola
For the hyperbola, the condition for a line \( y = mx + c \) to be a tangent is:
\[ c^2 = a^2 m^2 - b^2 \]
where \( a^2 = 16 \) and \( b^2 = 9 \). Thus, we have:
\[ c^2 = 16m^2 - 9 \]

### Step 4: Find the condition for the tangent to the ellipse
For the ellipse, the condition for a line \( y = mx + c \) to be a tangent is:
\[ c^2 = b^2 m^2 + a^2 \]
where \( a^2 = 4 \) and \( b^2 = 3 \). Thus, we have:
\[ c^2 = 3m^2 + 4 \]

### Step 5: Set the equations for \( c^2 \) equal to each other
Since the line is a common tangent to both curves, we can set the two expressions for \( c^2 \) equal:
\[ 16m^2 - 9 = 3m^2 + 4 \]

### Step 6: Solve for \( m^2 \)
Rearranging gives:
\[ 16m^2 - 3m^2 = 4 + 9 \]
\[ 13m^2 = 13 \]
\[ m^2 = 1 \]

### Step 7: Substitute \( m^2 \) back to find \( c^2 \)
Now substitute \( m^2 = 1 \) into either equation for \( c^2 \). Using \( c^2 = 16m^2 - 9 \):
\[ c^2 = 16(1) - 9 = 16 - 9 = 7 \]

### Step 8: Find \( m^2 + c^2 \)
Now we can find \( m^2 + c^2 \):
\[ m^2 + c^2 = 1 + 7 = 8 \]

### Final Answer
Thus, the value of \( m^2 + c^2 \) is:
\[ \boxed{8} \]