In `DeltaABC`, equation of side BC is x+y-6=0, also the circumcentre and orhtocentre are (3, 1) and (2, 2) respectively, then the equation of the circumcircle of `DeltaABC` is

Options

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

Explanation

To find the equation of the circumcircle of triangle ABC given the equation of side BC and the coordinates of the circumcenter and orthocenter, we can follow these steps:

### Step 1: Identify the given information
- The equation of side BC is \( x + y - 6 = 0 \).
- The circumcenter (O) is at (3, 1).
- The orthocenter (H) is at (2, 2).

### Step 2: Find the image of the orthocenter with respect to the circumcenter
The image of the orthocenter with respect to the circumcenter lies on the circumcircle. We can find the image (let's call it H') of the orthocenter (H) using the formula for reflection across a point.

Let the coordinates of the orthocenter be \( H(2, 2) \) and the circumcenter be \( O(3, 1) \).

Using the reflection formula:
\[ H' = O + (O - H) = (3, 1) + ((3, 1) - (2, 2)) \]
Calculating \( O - H \):
\[ (3 - 2, 1 - 2) = (1, -1) \]
Now, adding this vector to the circumcenter:
\[ H' = (3 + 1, 1 - 1) = (4, 0) \]

### Step 3: Calculate the radius of the circumcircle
The radius of the circumcircle can be calculated as the distance from the circumcenter (3, 1) to the point H' (4, 0).

Using the distance formula:
\[ r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates:
\[ r = \sqrt{(4 - 3)^2 + (0 - 1)^2} = \sqrt{1^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \]

### Step 4: Write the equation of the circumcircle
The general equation of a circle with center \((h, k)\) and radius \(r\) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Here, \(h = 3\), \(k = 1\), and \(r^2 = 2\).

Thus, the equation becomes:
\[ (x - 3)^2 + (y - 1)^2 = 2 \]

### Step 5: Expand the equation
Expanding the equation:
\[ (x^2 - 6x + 9) + (y^2 - 2y + 1) = 2 \]
Combining like terms:
\[ x^2 + y^2 - 6x - 2y + 10 = 2 \]
Subtracting 2 from both sides:
\[ x^2 + y^2 - 6x - 2y + 8 = 0 \]

### Final Equation
The final equation of the circumcircle of triangle ABC is:
\[ x^2 + y^2 - 6x - 2y + 8 = 0 \]