Find the equation of the circle which cuts the circle `x^(2)+y^(2)+5x+7y-4=0` orthogonally, has its centre on the line x=2 and passes through the point (4,-1).

Answer

To find the equation of the circle that cuts the given circle orthogonally, has its center on the line \( x = 2 \), and passes through the point \( (4, -1) \), we can follow these steps:

### Step 1: Identify the given circle's equation and parameters
The given circle is:
\[ x^2 + y^2 + 5x + 7y - 4 = 0 \]
We can rewrite this in the standard form \( (x + g_1)^2 + (y + f_1)^2 = r_1^2 \) to find its center and radius.

### Step 2: Complete the square
To find the center and radius, we complete the square for \( x \) and \( y \):

1. For \( x \):
\[ x^2 + 5x = (x + \frac{5}{2})^2 - \frac{25}{4} \]

2. For \( y \):
\[ y^2 + 7y = (y + \frac{7}{2})^2 - \frac{49}{4} \]

Putting it all together:
\[ (x + \frac{5}{2})^2 + (y + \frac{7}{2})^2 - \frac{25}{4} - \frac{49}{4} - 4 = 0 \]
\[ (x + \frac{5}{2})^2 + (y + \frac{7}{2})^2 = \frac{25 + 49 + 16}{4} = \frac{90}{4} = \frac{45}{2} \]

Thus, the center \( C_1 \) is \( (-\frac{5}{2}, -\frac{7}{2}) \) and the radius \( r_1 = \sqrt{\frac{45}{2}} \).

### Step 3: Set up the equation for the new circle
Let the new circle \( C_2 \) have its center at \( (2, f_2) \) and radius \( r_2 \). The general equation of the circle is:
\[ (x - 2)^2 + (y - f_2)^2 = r_2^2 \]

### Step 4: Use the orthogonality condition
For two circles to cut orthogonally, the following condition must hold:
\[ 2(g_1g_2 + f_1f_2) = c_1 + c_2 \]
Where \( g_1 = \frac{5}{2}, f_1 = \frac{7}{2}, c_1 = -4 \) for the first circle, and \( g_2 = -2, f_2, c_2 \) for the second circle.

Substituting these values:
\[ 2\left(\frac{5}{2} \cdot (-2) + \frac{7}{2} \cdot f_2\right) = -4 + c_2 \]
This simplifies to:
\[ -10 + 7f_2 = c_2 - 4 \]
Thus, we have:
\[ 7f_2 + 6 = c_2 \quad \text{(Equation 1)} \]

### Step 5: Use the point through which the circle passes
The circle \( C_2 \) passes through the point \( (4, -1) \). Substituting this point into the circle's equation:
\[ (4 - 2)^2 + (-1 - f_2)^2 = r_2^2 \]
This simplifies to:
\[ 4 + (f_2 + 1)^2 = r_2^2 \quad \text{(Equation 2)} \]

### Step 6: Express \( r_2^2 \) in terms of \( f_2 \)
From the general equation of the circle:
\[ r_2^2 = (2 + g_2)^2 + (f_2)^2 = 4 + 4 + f_2^2 = 8 + f_2^2 \]

### Step 7: Substitute \( r_2^2 \) into Equation 2
Now substituting \( r_2^2 \) into Equation 2:
\[ 4 + (f_2 + 1)^2 = 8 + f_2^2 \]
Expanding and simplifying:
\[ 4 + f_2^2 + 2f_2 + 1 = 8 + f_2^2 \]
This leads to:
\[ 2f_2 + 5 = 8 \Rightarrow 2f_2 = 3 \Rightarrow f_2 = \frac{3}{2} \]

### Step 8: Find \( c_2 \)
Substituting \( f_2 \) back into Equation 1:
\[ c_2 = 7\left(\frac{3}{2}\right) + 6 = \frac{21}{2} + 6 = \frac{21}{2} + \frac{12}{2} = \frac{33}{2} \]

### Step 9: Write the equation of the circle \( C_2 \)
Now substituting \( g_2 = -2, f_2 = \frac{3}{2}, c_2 = \frac{33}{2} \) into the circle equation:
\[ (x - 2)^2 + (y - \frac{3}{2})^2 = \frac{33}{2} \]
Expanding this gives:
\[ (x^2 - 4x + 4) + (y^2 - 3y + \frac{9}{4}) = \frac{33}{2} \]
Multiplying through by 4 to eliminate the fraction:
\[ 4x^2 - 16x + 16 + 4y^2 - 12y + 9 = 66 \]
Combining like terms:
\[ 4x^2 + 4y^2 - 16x - 12y - 41 = 0 \]
Dividing through by 4 gives the final equation:
\[ x^2 + y^2 - 4x - 3y - \frac{41}{4} = 0 \]

### Final Answer
The equation of the required circle is:
\[ x^2 + y^2 - 4x - 3y - \frac{41}{4} = 0 \]