The equation of two diameters of a cirlce are `x - y = 5` and `2 x + y = 4` and the radius of the circle is 5 units, then the equation of the circle is

Options

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

Answer

To find the equation of the circle given the diameters and radius, we can follow these steps:

### Step 1: Identify the equations of the diameters
The equations of the two diameters are given as:
1. \( x - y = 5 \) (Equation 1)
2. \( 2x + y = 4 \) (Equation 2)

### Step 2: Solve the equations to find the center of the circle
To find the center of the circle, we need to solve these two equations simultaneously.

From Equation 1:
\[ x - y = 5 \]
We can express \( y \) in terms of \( x \):
\[ y = x - 5 \] (Equation 3)

Now, substitute Equation 3 into Equation 2:
\[ 2x + (x - 5) = 4 \]
Combine like terms:
\[ 3x - 5 = 4 \]
Add 5 to both sides:
\[ 3x = 9 \]
Divide by 3:
\[ x = 3 \]

Now substitute \( x = 3 \) back into Equation 3 to find \( y \):
\[ y = 3 - 5 = -2 \]

Thus, the center of the circle is at the point \( (3, -2) \).

### Step 3: Use the radius to write the equation of the circle
The radius of the circle is given as \( r = 5 \). The standard form of the equation of a circle with center \( (h, k) \) and radius \( r \) is:
\[ (x - h)^2 + (y - k)^2 = r^2 \]

Substituting \( h = 3 \), \( k = -2 \), and \( r = 5 \):
\[ (x - 3)^2 + (y + 2)^2 = 5^2 \]
\[ (x - 3)^2 + (y + 2)^2 = 25 \]

### Step 4: Expand the equation
Now, we will expand the equation:
\[ (x - 3)(x - 3) + (y + 2)(y + 2) = 25 \]
\[ x^2 - 6x + 9 + y^2 + 4y + 4 = 25 \]
Combine like terms:
\[ x^2 + y^2 - 6x + 4y + 13 = 25 \]
Subtract 25 from both sides:
\[ x^2 + y^2 - 6x + 4y - 12 = 0 \]

### Final Answer
The equation of the circle is:
\[ x^2 + y^2 - 6x + 4y - 12 = 0 \]