`(6, 0), (0, 6) and (7, 7)` are the vertices of a triangle. The circle inscribed in the triangle has the equation

Options

A. `x^(2)+y^(2)-9x-9y+36=0`
B. `x^(2)+y^(2)+9x-9y+36=0`
C. `x^(2)+y^(2)+9x+9y-36=0`
D. `x^(2)+y^(2)+18x-18y+36=0`

Explanation

To find the equation of the circle inscribed in the triangle with vertices at \( A(6, 0) \), \( B(0, 6) \), and \( C(7, 7) \), we will follow these steps:

### Step 1: Calculate the lengths of the sides of the triangle
We will use the distance formula to find the lengths of the sides \( a \), \( b \), and \( c \) of the triangle.

- Length \( a \) (opposite vertex \( A \)):
\[ a = BC = \sqrt{(0 - 7)^2 + (6 - 7)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \]

- Length \( b \) (opposite vertex \( B \)):
\[ b = AC = \sqrt{(6 - 7)^2 + (0 - 7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

- Length \( c \) (opposite vertex \( C \)):
\[ c = AB = \sqrt{(6 - 0)^2 + (0 - 6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

### Step 2: Calculate the coordinates of the incenter
The incenter \( I \) of the triangle can be found using the formula:
\[ I_x = \frac{a x_1 + b x_2 + c x_3}{a + b + c}, \quad I_y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} \]
Substituting the values:
\[ I_x = \frac{(5\sqrt{2})(6) + (5\sqrt{2})(0) + (6\sqrt{2})(7)}{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} = \frac{30\sqrt{2} + 0 + 42\sqrt{2}}{16\sqrt{2}} = \frac{72\sqrt{2}}{16\sqrt{2}} = \frac{72}{16} = 4.5 \]
\[ I_y = \frac{(5\sqrt{2})(0) + (5\sqrt{2})(6) + (6\sqrt{2})(7)}{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} = \frac{0 + 30\sqrt{2} + 42\sqrt{2}}{16\sqrt{2}} = \frac{72\sqrt{2}}{16\sqrt{2}} = \frac{72}{16} = 4.5 \]
Thus, the incenter \( I \) is at \( \left(\frac{9}{2}, \frac{9}{2}\right) \).

### Step 3: Calculate the radius of the incircle
The radius \( r \) of the incircle can be calculated using the formula:
\[ r = \frac{A}{s} \]
where \( A \) is the area of the triangle and \( s \) is the semi-perimeter.

- Semi-perimeter \( s \):
\[ s = \frac{a + b + c}{2} = \frac{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}}{2} = \frac{16\sqrt{2}}{2} = 8\sqrt{2} \]

- Area \( A \) can be calculated using the formula:
\[ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ A = \frac{1}{2} \left| 6(6 - 7) + 0(7 - 0) + 7(0 - 6) \right| = \frac{1}{2} \left| 6(-1) + 0 + 7(-6) \right| = \frac{1}{2} \left| -6 - 42 \right| = \frac{1}{2} \times 48 = 24 \]

- Now calculate the radius:
\[ r = \frac{A}{s} = \frac{24}{8\sqrt{2}} = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2} \]

### Step 4: Write the equation of the incircle
The equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting \( h = \frac{9}{2} \), \( k = \frac{9}{2} \), and \( r = \frac{3\sqrt{2}}{2} \):
\[ \left(x - \frac{9}{2}\right)^2 + \left(y - \frac{9}{2}\right)^2 = \left(\frac{3\sqrt{2}}{2}\right)^2 \]
\[ \left(x - \frac{9}{2}\right)^2 + \left(y - \frac{9}{2}\right)^2 = \frac{9}{2} \]

### Final Equation
Expanding this gives:
\[ x^2 - 9x + \frac{81}{4} + y^2 - 9y + \frac{81}{4} = \frac{9}{2} \]
\[ x^2 + y^2 - 9x - 9y + \frac{162}{4} - \frac{18}{4} = 0 \]
\[ x^2 + y^2 - 9x - 9y + \frac{144}{4} = 0 \]
\[ x^2 + y^2 - 9x - 9y + 36 = 0 \]

### Conclusion
The equation of the inscribed circle is:
\[ x^2 + y^2 - 9x - 9y + 36 = 0 \]