If coordinates of the vertices A, B, C of a triangle ABC are (6, 0), (0, 6) and (7, 7) respectively, then centre of the circle touching AB externally and BC and CA produced is

Options

A. `(-3, -3)`
B. `(-4, -4)`
C. `(-2, -2)`
D. None of these

Answer

To find the center of the circle touching sides AB externally and sides BC and CA produced, we will follow these steps:

### Step 1: Identify the coordinates of the vertices
The coordinates of the vertices of triangle ABC are given as:
- A(6, 0)
- B(0, 6)
- C(7, 7)

### Step 2: Calculate the lengths of the sides of the triangle
We will calculate the lengths of sides AB, BC, and CA using the distance formula:
\[ \text{Distance} = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]

1. **Length of AB:**
\[ AB = \sqrt{(0 - 6)^2 + (6 - 0)^2} = \sqrt{(-6)^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \]

2. **Length of BC:**
\[ BC = \sqrt{(7 - 0)^2 + (7 - 6)^2} = \sqrt{(7)^2 + (1)^2} = \sqrt{49 + 1} = \sqrt{50} = 5\sqrt{2} \]

3. **Length of CA:**
\[ CA = \sqrt{(6 - 7)^2 + (0 - 7)^2} = \sqrt{(-1)^2 + (-7)^2} = \sqrt{1 + 49} = \sqrt{50} = 5\sqrt{2} \]

### Step 3: Assign lengths to variables
Let:
- \( a = BC = 5\sqrt{2} \)
- \( b = CA = 5\sqrt{2} \)
- \( c = AB = 6\sqrt{2} \)

### Step 4: Calculate the coordinates of the incenter
The coordinates of the incenter (I) can be calculated using the formula:
\[ I_x = \frac{a \cdot x_1 + b \cdot x_2 + c \cdot x_3}{a + b + c} \]
\[ I_y = \frac{a \cdot y_1 + b \cdot y_2 + c \cdot y_3}{a + b + c} \]

Substituting the values:
- \( x_1 = 6, y_1 = 0 \) (for A)
- \( x_2 = 0, y_2 = 6 \) (for B)
- \( x_3 = 7, y_3 = 7 \) (for C)

### Step 5: Calculate \( I_x \)
\[ I_x = \frac{(5\sqrt{2}) \cdot 6 + (5\sqrt{2}) \cdot 0 + (6\sqrt{2}) \cdot 7}{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} \]
\[ = \frac{30\sqrt{2} + 0 + 42\sqrt{2}}{16\sqrt{2}} = \frac{72\sqrt{2}}{16\sqrt{2}} = \frac{72}{16} = 4.5 \]

### Step 6: Calculate \( I_y \)
\[ I_y = \frac{(5\sqrt{2}) \cdot 0 + (5\sqrt{2}) \cdot 6 + (6\sqrt{2}) \cdot 7}{5\sqrt{2} + 5\sqrt{2} + 6\sqrt{2}} \]
\[ = \frac{0 + 30\sqrt{2} + 42\sqrt{2}}{16\sqrt{2}} = \frac{72\sqrt{2}}{16\sqrt{2}} = \frac{72}{16} = 4.5 \]

### Step 7: Conclusion
Thus, the coordinates of the center of the circle touching AB externally and BC and CA produced is:
\[ \text{Center} = (4.5, 4.5) \]