The locus of a point which moves such that the tangents from it to the two circles `x^(2)+y^(2)-5x-3=0` and `3x^(2)+3y^(2)+2x+4y-6=0` are equal, is given by

Options

A. `2x^(2) + 2y ^(2) +7x + 4y - 3=0`
B. `17x + 4y + 3=0`
C. ` 4x^(2) + 4y^(2) -3x + 4y -9=0`
D. `13x-4y +15 =0`

Answer

To find the locus of a point from which the tangents to the two given circles are equal, we will use the concept of the radical axis. Here’s a step-by-step solution:

### Step 1: Write the equations of the circles
The equations of the two circles given are:
1. \( C_1: x^2 + y^2 - 5x - 3 = 0 \)
2. \( C_2: 3x^2 + 3y^2 + 2x + 4y - 6 = 0 \)

### Step 2: Convert the second circle to standard form
To convert \( C_2 \) into standard form, we divide the entire equation by 3:
\[ C_2: x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2 = 0 \]

### Step 3: Set up the equations for the radical axis
The radical axis is found by equating the two circle equations. We can rewrite the equations as follows:
- From \( C_1 \): \( x^2 + y^2 - 5x - 3 = 0 \)
- From \( C_2 \): \( x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2 = 0 \)

### Step 4: Subtract the equations
Now we subtract the second equation from the first:
\[ (x^2 + y^2 - 5x - 3) - (x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2) = 0 \]
This simplifies to:
\[ -5x - 3 - \left(\frac{2}{3}x + \frac{4}{3}y - 2\right) = 0 \]

### Step 5: Simplify the equation
Distributing the negative sign:
\[ -5x - 3 - \frac{2}{3}x - \frac{4}{3}y + 2 = 0 \]
Combine like terms:
\[ -\left(5 + \frac{2}{3}\right)x - \frac{4}{3}y - 1 = 0 \]
To combine the coefficients of \( x \):
\[ -\left(\frac{15}{3} + \frac{2}{3}\right)x - \frac{4}{3}y - 1 = 0 \]
This gives:
\[ -\frac{17}{3}x - \frac{4}{3}y - 1 = 0 \]

### Step 6: Multiply through by -3 to eliminate fractions
Multiplying the entire equation by -3 gives:
\[ 17x + 4y + 3 = 0 \]

### Final Answer
The locus of the point from which the tangents to the two circles are equal is given by:
\[ \boxed{17x + 4y + 3 = 0} \]