The diameters of a circle are along 2x+y-7=0 and x+3y- 11=0. Then, the equation of this circle, which also passes through (5,7) is:

Options

A. `x^(2)+y^(2)-4x-6y-16=0`
B. `x^(2)+y^(2)-4x-6y-20=0`
C. `x^(2)+y^(2)-4x-6y-12=0`
D. `x^(2)+y^(2)+4x+6y-12=0`

Answer

To find the equation of the circle whose diameters are along the lines given by the equations \(2x + y - 7 = 0\) and \(x + 3y - 11 = 0\), and which also passes through the point \((5, 7)\), we can follow these steps:

### Step 1: Find the intersection point of the two lines.
We need to solve the two equations simultaneously to find the center of the circle.

1. The first equation is:
\[ 2x + y - 7 = 0 \quad \text{(Equation 1)} \]
2. The second equation is:
\[ x + 3y - 11 = 0 \quad \text{(Equation 2)} \]

To eliminate \(y\), we can multiply Equation 2 by 2:
\[ 2(x + 3y - 11) = 0 \implies 2x + 6y - 22 = 0 \quad \text{(Equation 3)} \]

### Step 2: Subtract Equation 1 from Equation 3.
Now we subtract Equation 1 from Equation 3:
\[ (2x + 6y - 22) - (2x + y - 7) = 0 \]
This simplifies to:
\[ 6y - y - 22 + 7 = 0 \implies 5y - 15 = 0 \implies 5y = 15 \implies y = 3 \]

### Step 3: Substitute \(y\) back to find \(x\).
Now substitute \(y = 3\) back into Equation 2 to find \(x\):
\[ x + 3(3) - 11 = 0 \implies x + 9 - 11 = 0 \implies x - 2 = 0 \implies x = 2 \]

### Step 4: Determine the center of the circle.
The center of the circle is at the point \((2, 3)\).

### Step 5: Write the general equation of the circle.
The general equation of a circle is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
where \((h, k)\) is the center and \(r\) is the radius. We can also express it in the form:
\[ x^2 + y^2 + 2gx + 2fy + c = 0 \]
where \(g = -h\) and \(f = -k\).

For our circle:
\[ h = 2, \quad k = 3 \implies g = -2, \quad f = -3 \]

Thus, the equation becomes:
\[ x^2 + y^2 - 4x - 6y + c = 0 \]

### Step 6: Use the point \((5, 7)\) to find \(c\).
Substituting the point \((5, 7)\) into the equation:
\[ 5^2 + 7^2 - 4(5) - 6(7) + c = 0 \]
Calculating this gives:
\[ 25 + 49 - 20 - 42 + c = 0 \implies 25 + 49 - 20 - 42 + c = 0 \implies 12 + c = 0 \implies c = -12 \]

### Step 7: Write the final equation of the circle.
Substituting \(c = -12\) back into the equation:
\[ x^2 + y^2 - 4x - 6y - 12 = 0 \]

Thus, the equation of the circle is:
\[ \boxed{x^2 + y^2 - 4x - 6y - 12 = 0} \]