Show that the line `x + y sqrt(3) = 4` touches the circles `x^(2) + y^(2) - 4x - 4 sqrt( 3) y +12 = 0 and x^(2) y^(2) = 4 ` at the same point . Also find the coordinate of the point

Answer

To show that the line \( x + y \sqrt{3} = 4 \) touches the circles \( x^2 + y^2 - 4x - 4\sqrt{3}y + 12 = 0 \) and \( x^2 + y^2 = 4 \) at the same point, we will follow these steps:

### Step 1: Rewrite the line equation
The line equation is given as:
\[ x + y \sqrt{3} = 4 \]
We can express \( y \) in terms of \( x \):
\[ y = \frac{4 - x}{\sqrt{3}} \]

### Step 2: Substitute \( y \) into the first circle's equation
The first circle's equation is:
\[ x^2 + y^2 - 4x - 4\sqrt{3}y + 12 = 0 \]
Substituting \( y = \frac{4 - x}{\sqrt{3}} \) into the circle's equation:
\[ x^2 + \left(\frac{4 - x}{\sqrt{3}}\right)^2 - 4x - 4\sqrt{3}\left(\frac{4 - x}{\sqrt{3}}\right) + 12 = 0 \]
Calculating \( y^2 \):
\[ y^2 = \frac{(4 - x)^2}{3} = \frac{16 - 8x + x^2}{3} \]
Now substituting \( y^2 \) into the circle's equation:
\[ x^2 + \frac{16 - 8x + x^2}{3} - 4x - 4(4 - x) + 12 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 3x^2 + (16 - 8x + x^2) - 12x - 12(4 - x) + 36 = 0 \]
Simplifying:
\[ 3x^2 + 16 - 8x + x^2 - 12x - 48 + 12x + 36 = 0 \]
Combining like terms:
\[ 4x^2 - 8x + 4 = 0 \]
Dividing through by 4:
\[ x^2 - 2x + 1 = 0 \]
Factoring:
\[ (x - 1)^2 = 0 \]
Thus, \( x = 1 \).

### Step 3: Find the corresponding \( y \) value
Substituting \( x = 1 \) back into the equation for \( y \):
\[ y = \frac{4 - 1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]
So, the point of tangency with the first circle is \( (1, \sqrt{3}) \).

### Step 4: Check the second circle's equation
The second circle's equation is:
\[ x^2 + y^2 = 4 \]
Substituting \( x = 1 \) and \( y = \sqrt{3} \):
\[ 1^2 + (\sqrt{3})^2 = 1 + 3 = 4 \]
This confirms that the point \( (1, \sqrt{3}) \) lies on the second circle as well.

### Conclusion
The line \( x + y \sqrt{3} = 4 \) touches both circles at the point \( (1, \sqrt{3}) \).