The equation of the circle inscribed in the triangle formed by line `x = 0, y = 0,3x - 4y+6=0` is ...........

Answer

To find the equation of the circle inscribed in the triangle formed by the lines \( x = 0 \), \( y = 0 \), and \( 3x - 4y + 6 = 0 \), we can follow these steps:

### Step 1: Identify the vertices of the triangle
The triangle is formed by the intersection of the lines:
1. \( x = 0 \) (the y-axis)
2. \( y = 0 \) (the x-axis)
3. \( 3x - 4y + 6 = 0 \)

To find the intersection points:
- The intersection of \( x = 0 \) and \( y = 0 \) is the origin \( (0, 0) \).
- To find the intersection of \( 3x - 4y + 6 = 0 \) with the axes:
- Set \( y = 0 \):
\[ 3x + 6 = 0 \implies x = -2 \quad \text{(Point A: } (-2, 0) \text{)} \]
- Set \( x = 0 \):
\[ -4y + 6 = 0 \implies y = \frac{3}{2} \quad \text{(Point B: } (0, \frac{3}{2}) \text{)} \]

Thus, the vertices of the triangle are:
- \( A(-2, 0) \)
- \( B(0, 0) \)
- \( C(0, \frac{3}{2}) \)

### Step 2: Find the lengths of the sides of the triangle
Using the distance formula, we can find the lengths of the sides:
- Length \( AB \):
\[ AB = \sqrt{((-2) - 0)^2 + (0 - 0)^2} = 2 \]
- Length \( AC \):
\[ AC = \sqrt{(0 - 0)^2 + \left(0 - \frac{3}{2}\right)^2} = \frac{3}{2} \]
- Length \( BC \):
\[ BC = \sqrt{((-2) - 0)^2 + \left(0 - \frac{3}{2}\right)^2} = \sqrt{4 + \frac{9}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2} \]

### Step 3: Calculate the semi-perimeter \( s \)
The semi-perimeter \( s \) is given by:
\[ s = \frac{AB + AC + BC}{2} = \frac{2 + \frac{3}{2} + \frac{5}{2}}{2} = \frac{2 + 4}{2} = 3 \]

### Step 4: Calculate the area \( A \) of the triangle
The area \( A \) can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times \frac{3}{2} = \frac{3}{2} \]

### Step 5: Find the radius \( r \) of the inscribed circle
The radius \( r \) of the inscribed circle is given by:
\[ r = \frac{A}{s} = \frac{\frac{3}{2}}{3} = \frac{1}{2} \]

### Step 6: Find the coordinates of the center of the inscribed circle
The center \( (h, k) \) of the inscribed circle can be found using the formula:
\[ h = \frac{aA_x + bB_x + cC_x}{a + b + c}, \quad k = \frac{aA_y + bB_y + cC_y}{a + b + c} \]
Where \( A_x, A_y \) are the coordinates of vertex A, and \( a, b, c \) are the lengths of the opposite sides:
- \( a = BC = \frac{5}{2} \)
- \( b = AC = \frac{3}{2} \)
- \( c = AB = 2 \)

Calculating \( h \) and \( k \):
\[ h = \frac{\frac{5}{2}(-2) + \frac{3}{2}(0) + 2(0)}{\frac{5}{2} + \frac{3}{2} + 2} = \frac{-5}{5} = -1 \]
\[ k = \frac{\frac{5}{2}(0) + \frac{3}{2}(0) + 2\left(\frac{3}{2}\right)}{\frac{5}{2} + \frac{3}{2} + 2} = \frac{3}{5} \]

### Step 7: Write the equation of the circle
The equation of a circle with center \( (h, k) \) and radius \( r \) is given by:
\[ (x - h)^2 + (y - k)^2 = r^2 \]
Substituting the values:
\[ (x + 1)^2 + \left(y - \frac{3}{5}\right)^2 = \left(\frac{1}{2}\right)^2 \]
Expanding this, we get:
\[ (x + 1)^2 + \left(y - \frac{3}{5}\right)^2 = \frac{1}{4} \]

### Final Equation
The final equation of the inscribed circle is:
\[ (x + 1)^2 + \left(y - \frac{3}{5}\right)^2 = \frac{1}{4} \]