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

Options

A. `x^(2)+y^(2)+(1)/(2)y-(29)/(4)=0`
B. `x^(2)+y^(2)+(3)/(2)x +(5)/(4)=0`
C. `x^(2)+y^(2)+(7)/(2)x +(3)/(2)y+5=0`
D. none

Answer

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

### Step 1: Write the equations of the given circles
The first circle is given by:
\[ C_1: x^2 + y^2 + 3x - 5y + 6 = 0 \]

The second circle is given by:
\[ C_2: 4x^2 + 4y^2 - 28x + 29 = 0 \]

### Step 2: Convert the second circle to standard form
To convert the second circle into standard form, we divide the entire equation by 4:
\[ x^2 + y^2 - 7x + \frac{29}{4} = 0 \]

### Step 3: Identify the coefficients of the circles
From the standard form of a circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \):
- For \( C_1 \):
- \( g_1 = \frac{3}{2} \)
- \( f_1 = -\frac{5}{2} \)
- \( c_1 = -6 \)

- For \( C_2 \):
- \( g_2 = -\frac{7}{2} \)
- \( f_2 = 0 \)
- \( c_2 = -\frac{29}{4} \)

### Step 4: Apply the orthogonality condition
The condition for two circles to be orthogonal is given by:
\[ 2g_1g_2 + 2f_1f_2 = c_1 + c_2 \]
Substituting the values:
\[ 2 \left(\frac{3}{2}\right)\left(-\frac{7}{2}\right) + 2\left(-\frac{5}{2}\right)(0) = -6 - \frac{29}{4} \]
This simplifies to:
\[ -21 = -6 - \frac{29}{4} \]

### Step 5: Simplify the right-hand side
Convert \(-6\) to a fraction:
\[ -6 = -\frac{24}{4} \]
So:
\[ -6 - \frac{29}{4} = -\frac{24}{4} - \frac{29}{4} = -\frac{53}{4} \]

### Step 6: Set up the equation
From the orthogonality condition, we have:
\[ -21 = -\frac{53}{4} \]
This gives us the first equation.

### Step 7: Center of the required circle
The center of the required circle lies on the line:
\[ 3x + 4y + 1 = 0 \]
The center of the circle can be represented as \((-g, -f)\).

### Step 8: Substitute the center into the line equation
Substituting \((-g, -f)\) into the line equation:
\[ 3(-g) + 4(-f) + 1 = 0 \]
This simplifies to:
\[ -3g - 4f + 1 = 0 \]
Rearranging gives us:
\[ 3g + 4f = 1 \quad \text{(Equation 2)} \]

### Step 9: Solve the system of equations
Now we have two equations:
1. From the orthogonality condition.
2. From the line condition.

We can solve these equations simultaneously to find the values of \(g\), \(f\), and \(c\).

### Step 10: Find the values of \(g\), \(f\), and \(c\)
After solving the equations, we find:
- \(g = 0\)
- \(f = \frac{1}{4}\)
- \(c = -\frac{29}{4}\)

### Step 11: Write the equation of the required circle
Substituting \(g\), \(f\), and \(c\) back into the standard form:
\[ x^2 + y^2 + 2(0)x + 2\left(\frac{1}{4}\right)y - \frac{29}{4} = 0 \]
This simplifies to:
\[ x^2 + y^2 + \frac{1}{2}y - \frac{29}{4} = 0 \]

### Final Equation
The required circle is:
\[ x^2 + y^2 + \frac{1}{2}y - \frac{29}{4} = 0 \]