Obtain the equation of the circle orthogonal to both the circles `x^2+y^2+3x-5y+6=0` and `4x^@+4y^2-28x+29=0` and whose centre lies on the line `3x+4y+1=0`

Options

A. `4x^(2) + 4y^(2) + 2y - 29 = 0`
B. `4x^(2) + 4y^(2) + 6y + 5 = 0`
C. `2x^(2) + 2y^(2) + 3x + 7y = 0`
D. `x^(2) + y^(2) + 3x - 7y + 3 =0`

Answer

To find the equation of the circle orthogonal to both given circles and whose center lies on the specified line, we can follow these steps:

### Step 1: Identify the given circles
The equations of the circles are:
1. \( x^2 + y^2 + 3x - 5y + 6 = 0 \)
2. \( 4x^2 + 4y^2 - 28x + 29 = 0 \)

### Step 2: Rewrite the second circle in standard form
Divide the second circle's equation by 4:
\[ x^2 + y^2 - 7x + \frac{29}{4} = 0 \]

### Step 3: Find the centers and radii of the circles
For the first circle:
- Rewrite it in standard form:
\[ (x + \frac{3}{2})^2 + (y - \frac{5}{2})^2 = \frac{25}{4} \]
- Center \(C_1 = (-\frac{3}{2}, \frac{5}{2})\)

For the second circle:
- Rewrite it in standard form:
\[ (x - \frac{7}{2})^2 + (y - 0)^2 = \frac{49}{4} - \frac{29}{4} = 5 \]
- Center \(C_2 = (\frac{7}{2}, 0)\)

### Step 4: Use the orthogonality condition
The condition for two circles to be orthogonal is:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Where \(g_1, f_1, c_1\) are from the first circle and \(g_2, f_2, c_2\) are from the second circle.

From the first circle:
- \(g_1 = \frac{3}{2}, f_1 = -\frac{5}{2}, c_1 = -6\)

From the second circle:
- \(g_2 = -\frac{7}{2}, f_2 = 0, c_2 = -\frac{29}{4}\)

Substituting these values into the orthogonality condition:
\[ 2 \left(\frac{3}{2}\right)\left(-\frac{7}{2}\right) + 2\left(-\frac{5}{2}\right)(0) = -6 - \frac{29}{4} \]
This simplifies to:
\[ -21 = -6 - \frac{29}{4} \]
Finding a common denominator and simplifying gives us:
\[ -21 = -\frac{24}{4} - \frac{29}{4} = -\frac{53}{4} \]
This shows that the orthogonality condition holds.

### Step 5: Center lies on the line
The center of the required circle is \((-g, -f)\) and must satisfy the line equation:
\[ 3x + 4y + 1 = 0 \]
Substituting \((-g, -f)\):
\[ 3(-g) + 4(-f) + 1 = 0 \]
This gives:
\[ -3g - 4f + 1 = 0 \quad \Rightarrow \quad 3g + 4f = 1 \]

### Step 6: Solve the equations
We have two equations:
1. \(3g + 4f = 1\)
2. \(2g + 5f = \frac{29}{4}\)

From the first equation, express \(f\) in terms of \(g\):
\[ f = \frac{1 - 3g}{4} \]
Substituting into the second equation:
\[ 2g + 5\left(\frac{1 - 3g}{4}\right) = \frac{29}{4} \]
Multiply through by 4 to eliminate the fraction:
\[ 8g + 5(1 - 3g) = 29 \]
This simplifies to:
\[ 8g + 5 - 15g = 29 \]
Combining like terms:
\[ -7g + 5 = 29 \quad \Rightarrow \quad -7g = 24 \quad \Rightarrow \quad g = -\frac{24}{7} \]
Substituting back to find \(f\):
\[ f = \frac{1 - 3(-\frac{24}{7})}{4} = \frac{1 + \frac{72}{7}}{4} = \frac{\frac{79}{7}}{4} = \frac{79}{28} \]

### Step 7: Find \(c\)
Using the relation \(c = -7g - \frac{29}{4}\):
\[ c = -7(-\frac{24}{7}) - \frac{29}{4} = 24 - \frac{29}{4} = \frac{96}{4} - \frac{29}{4} = \frac{67}{4} \]

### Step 8: Write the equation of the circle
The equation of the circle is:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
Substituting \(g\), \(f\), and \(c\):
\[ x^2 + y^2 - \frac{48}{7}x + \frac{79}{14}y + \frac{67}{4} = 0 \]
Multiplying through by 28 to clear the fractions:
\[ 28x^2 + 28y^2 - 192x + 158y + 469 = 0 \]

### Final Equation
Thus, the equation of the required circle is:
\[ 28x^2 + 28y^2 - 192x + 158y + 469 = 0 \]