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) and whose center lies on the line \(12x + 5y = 25\), we can follow these steps:

### Step 1: Write the standard equation of the circle
The standard equation of a circle is given by:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \((-g, -f)\) is the center of the circle.

### Step 2: Substitute point A(0, 5) into the equation
Since point A lies on the circle, 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:
\[ 25 + 10f + c = 0 \quad \text{(Equation 1)} \]

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

### Step 4: Use the condition that the center lies on the line
The center of the circle is \((-g, -f)\). Since the center lies on the line \(12x + 5y = 25\), we substitute:
\[ 12(-g) + 5(-f) = 25 \]
This simplifies to:
\[ -12g - 5f = 25 \quad \text{(Equation 3)} \]

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

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

Substituting \(c\) into Equation 2:
\[ 12g + 2f - 25 - 10f = -37 \]
This simplifies to:
\[ 12g - 8f = -12 \quad \text{(Equation 4)} \]

### Step 6: Substitute Equation 4 into Equation 3
From Equation 3:
\[ -12g - 5f = 25 \]
Now, we can express \(g\) from Equation 4:
\[ g = \frac{2f - 3}{3} \]
Substituting this into Equation 3:
\[ -12\left(\frac{2f - 3}{3}\right) - 5f = 25 \]
This simplifies to:
\[ -8f + 12 - 5f = 25 \]
Combining like terms:
\[ -13f + 12 = 25 \]
Thus:
\[ -13f = 13 \implies f = -1 \]

### Step 7: Find \(g\)
Substituting \(f = -1\) back into Equation 4:
\[ 12g - 8(-1) = -12 \]
This simplifies to:
\[ 12g + 8 = -12 \implies 12g = -20 \implies g = -\frac{5}{3} \]

### Step 8: Find \(c\)
Now substituting \(f = -1\) into the expression for \(c\):
\[ c = -25 - 10(-1) = -25 + 10 = -15 \]

### Step 9: Write the equation of the circle
Now we have \(g = -\frac{5}{3}\), \(f = -1\), and \(c = -15\). Substituting these values into the standard equation of the circle:
\[ x^2 + y^2 + 2\left(-\frac{5}{3}\right)x + 2(-1)y - 15 = 0 \]
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 Equation
Thus, the equation of the circle is:
\[ 3x^2 + 3y^2 - 10x - 6y - 45 = 0 \]