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

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 solve the problem, we need to find the locus of a point \( P(h, k) \) such that the lengths of the tangents drawn from this point to the two given circles are equal.

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

### Step 2: Rewrite the second circle in standard form
To simplify the second circle, we divide the entire equation by 3:
\[ x^2 + y^2 + \frac{2}{3}x + \frac{4}{3}y - 2 = 0 \]

### Step 3: Identify the center and radius of both circles
For the first circle:
- Completing the square:
\[ x^2 - 5x + y^2 - 3 = 0 \implies (x - \frac{5}{2})^2 + y^2 = \frac{25}{4} + 3 = \frac{37}{4} \]
Center: \( \left(\frac{5}{2}, 0\right) \), Radius: \( \frac{\sqrt{37}}{2} \)

For the second circle:
- Completing the square:
\[ x^2 + \frac{2}{3}x + y^2 + \frac{4}{3}y - 2 = 0 \]
\[ \left(x + \frac{1}{3}\right)^2 + \left(y + \frac{2}{3}\right)^2 = 2 + \frac{1}{9} + \frac{4}{9} = \frac{25}{9} \]
Center: \( \left(-\frac{1}{3}, -\frac{2}{3}\right) \), Radius: \( \frac{5}{3} \)

### Step 4: Write the formula for the length of the tangent
The length of the tangent from a point \( P(h, k) \) to a circle defined by \( x^2 + y^2 + Dx + Ey + F = 0 \) is given by:
\[ L = \sqrt{h^2 + k^2 + Dh + Ek + F} \]

### Step 5: Calculate the lengths of the tangents
For the first circle:
\[ L_1 = \sqrt{h^2 + k^2 - 5h - 3} \]

For the second circle:
\[ L_2 = \sqrt{h^2 + k^2 + \frac{2}{3}h + \frac{4}{3}k - 2} \]

### Step 6: Set the lengths equal
Since the lengths of the tangents are equal:
\[ L_1 = L_2 \]
Squaring both sides:
\[ h^2 + k^2 - 5h - 3 = h^2 + k^2 + \frac{2}{3}h + \frac{4}{3}k - 2 \]

### Step 7: Simplify the equation
Cancelling \( h^2 + k^2 \) from both sides:
\[ -5h - 3 = \frac{2}{3}h + \frac{4}{3}k - 2 \]
Rearranging gives:
\[ -5h - \frac{2}{3}h - \frac{4}{3}k + 2 - 3 = 0 \]
Multiplying through by 3 to eliminate fractions:
\[ -15h - 2h - 4k + 6 - 9 = 0 \]
Combining terms:
\[ -17h - 4k - 3 = 0 \]
Thus:
\[ 17h + 4k + 3 = 0 \]

### Step 8: Write the locus equation
Substituting \( h \) and \( k \) back to \( x \) and \( y \):
\[ 17x + 4y + 3 = 0 \]

### Final Answer
The locus of the point \( P(h, k) \) is given by the equation:
\[ 17x + 4y + 3 = 0 \]