The equation of the circle inscribed in the triangle formed by the coordinate axes and the lines ` 12x+ 5y = 60` is `:`

Options

A. `x^(2) +y^(2) +4x + 4y+4 =0`
B. ` x^(2) + y^(2) -4x -4y+4 =0`
C. ` x^(2) +y^(2) - 4x -4y -4=0`
D. None of these

Answer

To find the equation of the circle inscribed in the triangle formed by the coordinate axes and the line \(12x + 5y = 60\), we can follow these steps:

### Step 1: Identify the intercepts of the line
The line \(12x + 5y = 60\) intersects the x-axis and y-axis. To find these intercepts:
- **X-intercept**: Set \(y = 0\):
\[ 12x = 60 \implies x = 5 \]
- **Y-intercept**: Set \(x = 0\):
\[ 5y = 60 \implies y = 12 \]

Thus, the triangle is formed by the points \((0, 0)\), \((5, 0)\), and \((0, 12)\).

### Step 2: Determine the coordinates of the incenter
The incenter of a triangle formed by the coordinate axes and a line can be found using the formula for the incenter coordinates:
\[ (h, k) = \left( \frac{a}{a+b+c} \cdot A_x, \frac{b}{a+b+c} \cdot A_y \right) \]
where \(A_x\) and \(A_y\) are the x and y coordinates of the vertices, and \(a\), \(b\), and \(c\) are the lengths of the sides opposite to these vertices.

In our case:
- The lengths of the sides are:
- \(a = 12\) (opposite to vertex \((5, 0)\))
- \(b = 5\) (opposite to vertex \((0, 12)\))
- \(c = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\) (the hypotenuse)

### Step 3: Calculate the incenter coordinates
Using the formula:
\[ h = \frac{a}{a+b+c} \cdot 5 + \frac{b}{a+b+c} \cdot 0 = \frac{12}{12+5+13} \cdot 5 = \frac{12 \cdot 5}{30} = 2 \]
\[ k = \frac{b}{a+b+c} \cdot 12 + \frac{a}{a+b+c} \cdot 0 = \frac{5}{30} \cdot 12 = 2 \]

Thus, the incenter is at the point \((2, 2)\).

### Step 4: Determine the radius of the inscribed circle
The radius \(r\) of the inscribed circle is equal to the distance from the incenter to any side of the triangle. Since the incenter is \((2, 2)\), we can calculate the distance from this point to the line \(12x + 5y - 60 = 0\) using the formula:
\[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
where \(A = 12\), \(B = 5\), \(C = -60\), and \((x_0, y_0) = (2, 2)\):
\[ \text{Distance} = \frac{|12 \cdot 2 + 5 \cdot 2 - 60|}{\sqrt{12^2 + 5^2}} = \frac{|24 + 10 - 60|}{\sqrt{144 + 25}} = \frac{|-26|}{13} = 2 \]

### Step 5: Write the equation of the inscribed 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 \(h = 2\), \(k = 2\), and \(r = 2\):
\[ (x - 2)^2 + (y - 2)^2 = 2^2 \]
This simplifies to:
\[ (x - 2)^2 + (y - 2)^2 = 4 \]

### Final Answer
The equation of the circle inscribed in the triangle is:
\[ (x - 2)^2 + (y - 2)^2 = 4 \]