if `y=Mx+C`is a common tangent of circle `x^2+y^2= 3` hyperbola `x^2/64-y^2/100 =1` then which of the following statement is true.

Options

A. `8M=4`
B. `61C^2=492`
C. `4C^2=369`
D. `8M+5=0`

Answer

To solve the problem, we need to find the conditions under which the line \( y = Mx + C \) is a common tangent to both the circle \( x^2 + y^2 = 3 \) and the hyperbola \( \frac{x^2}{64} - \frac{y^2}{100} = 1 \).

### Step 1: Find the relation for the circle
The equation of the circle is given by \( x^2 + y^2 = 3 \). The condition for a line \( y = mx + c \) to be a tangent to the circle can be derived from the formula:
\[ c^2 = r^2(1 + m^2) \]
where \( r \) is the radius of the circle. Here, \( r^2 = 3 \), so we have:
\[ c^2 = 3(1 + m^2) \tag{1} \]

### Step 2: Find the relation for the hyperbola
The equation of the hyperbola is given by \( \frac{x^2}{64} - \frac{y^2}{100} = 1 \). The condition for a line \( y = mx + c \) to be a tangent to the hyperbola is:
\[ c^2 = a^2 m^2 - b^2 \]
where \( a^2 = 64 \) and \( b^2 = 100 \). Therefore, we have:
\[ c^2 = 64m^2 - 100 \tag{2} \]

### Step 3: Set the two equations for \( c^2 \) equal to each other
Since both equations represent \( c^2 \), we can set them equal:
\[ 3(1 + m^2) = 64m^2 - 100 \]

### Step 4: Simplify the equation
Expanding and rearranging gives:
\[ 3 + 3m^2 = 64m^2 - 100 \]
\[ 3 + 100 = 64m^2 - 3m^2 \]
\[ 103 = 61m^2 \]
\[ m^2 = \frac{103}{61} \]

### Step 5: Find \( c^2 \)
Now that we have \( m^2 \), we can substitute it back into either equation (1) or (2) to find \( c^2 \). We will use equation (1):
\[ c^2 = 3(1 + \frac{103}{61}) = 3\left(\frac{61 + 103}{61}\right) = 3\left(\frac{164}{61}\right) = \frac{492}{61} \]

### Conclusion
Thus, the values we have derived are:
- \( m^2 = \frac{103}{61} \)
- \( c^2 = \frac{492}{61} \)

Now we can check which of the given statements in the options is true based on these results.