Find the coordinates of the point from which the lengths of the tangents to the following three circles be equal `3x^(2)+3y^(2)+4x-6y-1=0,2x^(2)+2y^(2)-3x-2y-4=0and2x^(2)+2y^(2)-x+y-1=0`

Answer

To find the coordinates of the point from which the lengths of the tangents to the given three circles are equal, we will follow these steps:

### Step 1: Write the equations of the circles in standard form.

1. **Circle 1**:
\[ 3x^2 + 3y^2 + 4x - 6y - 1 = 0 \]
Divide by 3:
\[ x^2 + y^2 + \frac{4}{3}x - 2y - \frac{1}{3} = 0 \]

2. **Circle 2**:
\[ 2x^2 + 2y^2 - 3x - 2y - 4 = 0 \]
Divide by 2:
\[ x^2 + y^2 - \frac{3}{2}x - y - 2 = 0 \]

3. **Circle 3**:
\[ 2x^2 + 2y^2 - x + y - 1 = 0 \]
Divide by 2:
\[ x^2 + y^2 - \frac{1}{2}x + \frac{1}{2}y - \frac{1}{2} = 0 \]

### Step 2: Set up the equations for the radical axis.

The radical axis of two circles can be found using the equation \( S_1 - S_2 = 0 \).

1. **Radical Axis of Circle 1 and Circle 2**:
\[ S_1 - S_2 = 0 \]
\[ \left(x^2 + y^2 + \frac{4}{3}x - 2y - \frac{1}{3}\right) - \left(x^2 + y^2 - \frac{3}{2}x - y - 2\right) = 0 \]
Simplifying gives:
\[ \frac{17}{6}x - y + \frac{5}{3} = 0 \quad \text{(Equation 1)} \]

2. **Radical Axis of Circle 2 and Circle 3**:
\[ S_2 - S_3 = 0 \]
\[ \left(x^2 + y^2 - \frac{3}{2}x - y - 2\right) - \left(x^2 + y^2 - \frac{1}{2}x + \frac{1}{2}y - \frac{1}{2}\right) = 0 \]
Simplifying gives:
\[ -x - \frac{3}{2}y - \frac{3}{2} = 0 \quad \text{(Equation 2)} \]

### Step 3: Solve the system of equations.

Now we will solve the two equations obtained from the radical axes.

1. From Equation 1:
\[ y = \frac{17}{6}x + \frac{5}{3} \]

2. Substitute \( y \) in Equation 2:
\[ -x - \frac{3}{2}\left(\frac{17}{6}x + \frac{5}{3}\right) - \frac{3}{2} = 0 \]
Simplifying:
\[ -x - \frac{51}{12}x - \frac{15}{6} - \frac{3}{2} = 0 \]
Combine like terms and solve for \( x \):
\[ -\left(1 + \frac{51}{12}\right)x - \frac{15 + 18}{12} = 0 \]
\[ -\frac{63}{12}x - \frac{33}{12} = 0 \]
\[ x = -\frac{33}{63} = -\frac{11}{21} \]

3. Substitute \( x \) back to find \( y \):
\[ y = \frac{17}{6}\left(-\frac{11}{21}\right) + \frac{5}{3} \]
\[ y = -\frac{187}{126} + \frac{70}{126} = -\frac{117}{126} = -\frac{39}{42} \]

### Final Coordinates
Thus, the coordinates of the point from which the lengths of the tangents to the three circles are equal are:
\[ \left(-\frac{11}{21}, -\frac{39}{42}\right) \]