The equation of the circle which passes through the point A(0, 5) and B(6, 1) and whose centre lies on the line `12x+5y=25` is

Options

A. `3x^(2)+3y^(2)+10x+6y+15=0`
B. `3x^(2)+3y^(2)-10x-6y-45=0`
C. `x^(2)+y^(2)-6x-6y+5=0`
D. `x^(2)+y^(2)-4x-3y-10=0`

Answer

To find the equation of the circle that passes through the points A(0, 5) and B(6, 1) with its center lying on the line \(12x + 5y = 25\), we can follow these steps:

### Step 1: General Equation of the Circle
The general equation of a circle can be expressed as:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((-g, -f)\) is the center of the circle.

### Step 2: Substitute Point A(0, 5)
Since the circle passes through point A(0, 5), we substitute \(x = 0\) and \(y = 5\) into the circle's equation:
\[ 0^2 + 5^2 + 2g(0) + 2f(5) + c = 0 \]
This simplifies to:
\[ 25 + 10f + c = 0 \]
Let’s denote this as Equation (1):
\[ 10f + c = -25 \quad \text{(1)} \]

### Step 3: Substitute Point B(6, 1)
Next, we substitute point B(6, 1) into the circle's equation:
\[ 6^2 + 1^2 + 2g(6) + 2f(1) + c = 0 \]
This simplifies to:
\[ 36 + 1 + 12g + 2f + c = 0 \]
Thus, we have:
\[ 37 + 12g + 2f + c = 0 \]
Let’s denote this as Equation (2):
\[ 12g + 2f + c = -37 \quad \text{(2)} \]

### Step 4: Center Lies on the Line
The center of the circle \((-g, -f)\) lies on the line \(12x + 5y = 25\). Substituting \(-g\) and \(-f\) into the line equation gives:
\[ 12(-g) + 5(-f) = 25 \]
This simplifies to:
\[ -12g - 5f = 25 \]
Let’s denote this as Equation (3):
\[ 12g + 5f = -25 \quad \text{(3)} \]

### Step 5: Solve the System of Equations
Now we have a system of three equations:
1. \(10f + c = -25\) (1)
2. \(12g + 2f + c = -37\) (2)
3. \(12g + 5f = -25\) (3)

From Equation (1), we can express \(c\) in terms of \(f\):
\[ c = -25 - 10f \quad \text{(4)} \]

Substituting Equation (4) into Equation (2):
\[ 12g + 2f + (-25 - 10f) = -37 \]
This simplifies to:
\[ 12g - 8f - 25 = -37 \]
Rearranging gives:
\[ 12g - 8f = -12 \]
Dividing through by 4:
\[ 3g - 2f = -3 \quad \text{(5)} \]

Now we have Equations (3) and (5):
1. \(12g + 5f = -25\) (3)
2. \(3g - 2f = -3\) (5)

### Step 6: Solve for \(g\) and \(f\)
From Equation (5):
\[ 3g = 2f - 3 \Rightarrow g = \frac{2f - 3}{3} \quad \text{(6)} \]

Substituting Equation (6) into Equation (3):
\[ 12\left(\frac{2f - 3}{3}\right) + 5f = -25 \]
This simplifies to:
\[ 4(2f - 3) + 5f = -25 \]
Expanding gives:
\[ 8f - 12 + 5f = -25 \]
Combining like terms:
\[ 13f - 12 = -25 \]
Thus:
\[ 13f = -13 \Rightarrow f = -1 \]

### Step 7: Substitute \(f\) to Find \(g\)
Substituting \(f = -1\) back into Equation (6):
\[ g = \frac{2(-1) - 3}{3} = \frac{-2 - 3}{3} = \frac{-5}{3} \]

### Step 8: Substitute \(g\) and \(f\) to Find \(c\)
Now substituting \(f = -1\) into Equation (4) to find \(c\):
\[ c = -25 - 10(-1) = -25 + 10 = -15 \]

### Step 9: Write the Circle's Equation
Now we have \(g = -\frac{5}{3}\), \(f = -1\), and \(c = -15\). The equation of the circle becomes:
\[ x^2 + y^2 + 2\left(-\frac{5}{3}\right)x + 2(-1)y - 15 = 0 \]
This simplifies to:
\[ x^2 + y^2 - \frac{10}{3}x - 2y - 15 = 0 \]
Multiplying through by 3 to eliminate the fraction:
\[ 3x^2 + 3y^2 - 10x - 6y - 45 = 0 \]

### Final Answer
The equation of the circle is:
\[ 3x^2 + 3y^2 - 10x - 6y - 45 = 0 \]