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 solve the problem, we need to find the value of \( m^2 + c^2 \) for the common tangent \( y = mx + c \) to the hyperbola and the ellipse given by the equations:

1. Hyperbola: \( \frac{x^2}{16} - \frac{y^2}{9} = 1 \)
2. Ellipse: \( \frac{x^2}{4} + \frac{y^2}{3} = 1 \)

### Step 1: Write the equations of the tangents

For the hyperbola, the equation of the tangent line with slope \( m \) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 - b^2} \]
where \( a^2 = 16 \) and \( b^2 = 9 \). Thus, \( a = 4 \) and \( b = 3 \).

The tangent equation becomes:
\[ y = mx \pm \sqrt{16m^2 - 9} \]

For the ellipse, the equation of the tangent line with slope \( m \) is given by:
\[ y = mx \pm \sqrt{a^2 m^2 + b^2} \]
where \( a^2 = 4 \) and \( b^2 = 3 \). Thus, \( a = 2 \) and \( b = \sqrt{3} \).

The tangent equation becomes:
\[ y = mx \pm \sqrt{4m^2 + 3} \]

### Step 2: Set the two tangent equations equal

Since both equations represent the same line (common tangent), we can equate the intercepts:
\[ \sqrt{16m^2 - 9} = \sqrt{4m^2 + 3} \]

### Step 3: Square both sides to eliminate the square roots

Squaring both sides gives:
\[ 16m^2 - 9 = 4m^2 + 3 \]

### Step 4: Rearrange the equation

Rearranging the equation:
\[ 16m^2 - 4m^2 = 3 + 9 \]
\[ 12m^2 = 12 \]
\[ m^2 = 1 \]

### Step 5: Find the value of \( m \)

Taking the square root of both sides, we find:
\[ m = \pm 1 \]

### Step 6: Substitute \( m \) back to find \( c \)

Now, substituting \( m = 1 \) into the tangent equation for the hyperbola:
\[ y = 1 \cdot x \pm \sqrt{16(1)^2 - 9} \]
\[ y = x \pm \sqrt{16 - 9} \]
\[ y = x \pm \sqrt{7} \]

Thus, the intercept \( c \) can be \( \pm \sqrt{7} \).

### Step 7: Calculate \( m^2 + c^2 \)

Now we compute \( m^2 + c^2 \):
\[ m^2 + c^2 = 1^2 + (\sqrt{7})^2 = 1 + 7 = 8 \]

### Final Answer

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