From a point `P`, tangents drawn to the circle `x^2 + y^2 + x-3=0, 3x^2 + 3y^2 - 5x+3y=0 and 4x^2 + 4y^2 + 8x+7y+9=0` are of equal lengths. Find the equation of the circle through `P`, which touches the line `x+y=5` at the point `(6, -1)`.

Answer

To solve the problem step by step, we need to follow these steps:

### Step 1: Identify the circles and their centers
The given circles are:
1. \( x^2 + y^2 + x - 3 = 0 \)
2. \( 3x^2 + 3y^2 - 5x + 3y = 0 \)
3. \( 4x^2 + 4y^2 + 8x + 7y + 9 = 0 \)

We can rewrite these equations in standard form to identify their centers and radii.

1. For the first circle:
\[ x^2 + y^2 + x - 3 = 0 \implies (x + \frac{1}{2})^2 + y^2 = \frac{13}{4} \]
Center: \( C_1(-\frac{1}{2}, 0) \), Radius: \( \frac{\sqrt{13}}{2} \)

2. For the second circle:
\[ 3(x^2 + y^2) - 5x + 3y = 0 \implies (x - \frac{5}{6})^2 + (y + \frac{1}{2})^2 = \frac{25}{36} + \frac{1}{4} \]
Center: \( C_2(\frac{5}{6}, -\frac{1}{2}) \), Radius: \( \frac{5}{6} \)

3. For the third circle:
\[ 4(x^2 + y^2) + 8x + 7y + 9 = 0 \implies (x + 1)^2 + (y + \frac{7}{8})^2 = \frac{25}{16} \]
Center: \( C_3(-1, -\frac{7}{8}) \), Radius: \( \frac{5}{4} \)

### Step 2: Set up the condition for equal tangents
Let \( P(x_1, y_1) \) be the point from which tangents are drawn to these circles. The lengths of the tangents from point \( P \) to each circle must be equal.

The length of the tangent from point \( P \) to a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ L = \sqrt{(x_1 - h)^2 + (y_1 - k)^2 - r^2} \]

Setting the lengths equal for the three circles will give us a system of equations.

### Step 3: Find the equation of the circle through P
We need to find the equation of a circle that passes through point \( P \) and touches the line \( x + y = 5 \) at the point \( (6, -1) \).

The general equation of a circle can be written as:
\[ (x - 6)^2 + (y + 1)^2 + \lambda (x + y - 5) = 0 \]

### Step 4: Substitute the point through which the circle passes
The circle passes through the point \( (0, -3) \):
\[ (0 - 6)^2 + (-3 + 1)^2 + \lambda (0 + (-3) - 5) = 0 \]
Calculating this gives:
\[ 36 + 4 + \lambda (-8) = 0 \implies 40 - 8\lambda = 0 \implies \lambda = 5 \]

### Step 5: Substitute \( \lambda \) back into the equation
Now substitute \( \lambda = 5 \) back into the circle equation:
\[ (x - 6)^2 + (y + 1)^2 + 5(x + y - 5) = 0 \]

Expanding this:
\[ (x - 6)^2 + (y + 1)^2 + 5x + 5y - 25 = 0 \]
\[ x^2 - 12x + 36 + y^2 + 2y + 1 + 5x + 5y - 25 = 0 \]
Combining like terms:
\[ x^2 + y^2 - 7x + 7y + 12 = 0 \]

### Final Answer
The equation of the required circle is:
\[ x^2 + y^2 - 7x + 7y + 12 = 0 \]