Find the lengths of the tangents drawn
to the circle ` x^(2) + y^(2) + 2 bx - 3b ^(2) = 0` from the point (a + b, a - b)

Answer

To find the lengths of the tangents drawn to the circle given by the equation \( x^2 + y^2 + 2bx - 3b^2 = 0 \) from the point \( (a + b, a - b) \), we can follow these steps:

### Step 1: Rewrite the Circle Equation
The given circle equation is:
\[ x^2 + y^2 + 2bx - 3b^2 = 0 \]
We can rearrange this into the standard form of a circle:
\[ x^2 + y^2 + 2bx = 3b^2 \]
This can be rewritten as:
\[ (x + b)^2 + y^2 = 4b^2 \]
This shows that the center of the circle is at \( (-b, 0) \) and the radius \( r \) is \( 2b \).

### Step 2: Identify the External Point
The point from which we are drawing the tangents is \( (a + b, a - b) \).

### Step 3: Use the Length of Tangent Formula
The length of the tangent \( L \) from an external point \( (x_1, y_1) \) to a circle defined by \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is given by:
\[ L = \sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c} \]
In our case, we have:
- \( g = b \)
- \( f = 0 \)
- \( c = -3b^2 \)
- \( x_1 = a + b \)
- \( y_1 = a - b \)

### Step 4: Substitute the Values
Now we substitute these values into the formula:
\[ L = \sqrt{(a + b)^2 + (a - b)^2 + 2b(a + b) + 2 \cdot 0 \cdot (a - b) - 3b^2} \]

### Step 5: Simplify the Expression
Calculating each term:
1. \( (a + b)^2 = a^2 + 2ab + b^2 \)
2. \( (a - b)^2 = a^2 - 2ab + b^2 \)
3. \( 2b(a + b) = 2ab + 2b^2 \)

Putting it all together:
\[ L = \sqrt{(a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) + (2ab + 2b^2) - 3b^2} \]
This simplifies to:
\[ L = \sqrt{2a^2 + 2b^2 + 2ab + 2b^2 - 3b^2} \]
\[ L = \sqrt{2a^2 + 2b^2 + 2ab - b^2} \]
\[ L = \sqrt{2a^2 + b^2 + 2ab} \]

### Final Answer
Thus, the length of the tangents drawn to the circle from the point \( (a + b, a - b) \) is:
\[ L = \sqrt{2a^2 + 2ab + b^2} \]