If the length of the tangent drawn from `(f, g)` to the circle `x^2+y^2= 6` be twice the length of the tangent drawn from the same point to the circle `x^2 + y^2 + 3 (x + y) = 0` then show that `g^2 +f^2 + 4g + 4f+ 2 = 0 .`

Answer

To solve the problem, we need to find the relationship between the point \((f, g)\) and the two circles given.

### Step 1: Identify the circles and their equations
The first circle is given by:
\[ x^2 + y^2 = 6 \]
The second circle is given by:
\[ x^2 + y^2 + 3(x + y) = 0 \]
We can rewrite the second circle's equation as:
\[ x^2 + y^2 + 3x + 3y = 0 \]

### Step 2: Calculate the length of the tangent from the point \((f, g)\) to the first circle
The formula for the length of the tangent from a point \((x_1, y_1)\) to a circle \(x^2 + y^2 = r^2\) is given by:
\[ \text{Length} = \sqrt{x_1^2 + y_1^2 - r^2} \]
For the first circle, \(r^2 = 6\):
\[ \text{Length}_1 = \sqrt{f^2 + g^2 - 6} \]

### Step 3: Calculate the length of the tangent from the point \((f, g)\) to the second circle
For the second circle, we can rewrite the equation as:
\[ x^2 + y^2 = -3(x + y) \]
To find the radius, we can complete the square:
\[ (x + \frac{3}{2})^2 + (y + \frac{3}{2})^2 = \frac{9}{2} \]
This shows that the center of the circle is \((-1.5, -1.5)\) and the radius is \(\sqrt{\frac{9}{2}} = \frac{3}{\sqrt{2}}\).

Using the tangent length formula again:
\[ \text{Length}_2 = \sqrt{f^2 + g^2 + 3f + 3g} \]

### Step 4: Set up the equation based on the problem statement
According to the problem, the length of the tangent to the first circle is twice the length of the tangent to the second circle:
\[ \sqrt{f^2 + g^2 - 6} = 2 \sqrt{f^2 + g^2 + 3f + 3g} \]

### Step 5: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ f^2 + g^2 - 6 = 4(f^2 + g^2 + 3f + 3g) \]

### Step 6: Expand and rearrange the equation
Expanding the right side:
\[ f^2 + g^2 - 6 = 4f^2 + 4g^2 + 12f + 12g \]
Rearranging gives:
\[ f^2 + g^2 - 4f^2 - 4g^2 - 12f - 12g - 6 = 0 \]
This simplifies to:
\[ -3f^2 - 3g^2 - 12f - 12g - 6 = 0 \]
Dividing through by -3:
\[ f^2 + g^2 + 4f + 4g + 2 = 0 \]

### Conclusion
We have shown that:
\[ g^2 + f^2 + 4g + 4f + 2 = 0 \]