If the lines `x + y =6 and x + 2y =4` be diameters of the circel whose diameter is 20, then the equation of the circle is

Options

A. `x ^(2) + y^(2) -16x + 4y -32=0`
B. `x ^(2) + y^(2) + 16 x + 4y -32=0`
C. ` x ^(2) + y^(2) + 16 x + 4y + 32=0`
D. ` x ^(2) + y^(2) + 16x -4y + 32 =0`

Answer

To find the equation of the circle whose diameters are given by the lines \(x + y = 6\) and \(x + 2y = 4\), and whose diameter is 20, we can follow these steps:

### Step 1: Find the intersection point of the lines
To find the center of the circle, we need to determine the point where the two lines intersect. We can solve the equations:

1. \(x + y = 6\) (Equation 1)
2. \(x + 2y = 4\) (Equation 2)

We can express \(y\) from Equation 1:
\[ y = 6 - x \]

Now, substitute \(y\) in Equation 2:
\[ x + 2(6 - x) = 4 \]
\[ x + 12 - 2x = 4 \]
\[ - x + 12 = 4 \]
\[ - x = 4 - 12 \]
\[ - x = -8 \implies x = 8 \]

Now substitute \(x = 8\) back into Equation 1 to find \(y\):
\[ 8 + y = 6 \implies y = 6 - 8 = -2 \]

Thus, the center of the circle is at the point \((8, -2)\).

### Step 2: Determine the radius of the circle
The diameter of the circle is given as 20, so the radius \(r\) is:
\[ r = \frac{diameter}{2} = \frac{20}{2} = 10 \]

### Step 3: Write the standard equation of the circle
The standard equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \(h = 8\), \(k = -2\), and \(r = 10\):
\[ (x - 8)^2 + (y + 2)^2 = 10^2 \]
\[ (x - 8)^2 + (y + 2)^2 = 100 \]

### Step 4: Expand the equation
Now we will expand the equation:
\[ (x - 8)^2 + (y + 2)^2 = 100 \]
Expanding both squares:
\[ (x^2 - 16x + 64) + (y^2 + 4y + 4) = 100 \]
Combining like terms:
\[ x^2 + y^2 - 16x + 4y + 68 = 100 \]
Rearranging gives:
\[ x^2 + y^2 - 16x + 4y + 68 - 100 = 0 \]
\[ x^2 + y^2 - 16x + 4y - 32 = 0 \]

### Final Equation
Thus, the equation of the circle is:
\[ x^2 + y^2 - 16x + 4y - 32 = 0 \]