For a `DeltaABC` the vertices are `A(0, 3), B(0, 12) and C(x, 0)`. If the circumcircle of `DeltaABC` touches the x - axis, then the area (in sq. units) of the `DeltaABC` is

Options

A. 36
B. 27
C. 30
D. 24

Answer

To find the area of triangle \( \Delta ABC \) with vertices \( A(0, 3) \), \( B(0, 12) \), and \( C(x, 0) \), given that the circumcircle touches the x-axis, we can follow these steps:

### Step 1: Understand the Geometry
The circumcircle of triangle \( ABC \) touches the x-axis. This means the radius of the circumcircle is equal to the distance from the center of the circle to the x-axis.

### Step 2: Determine the Center and Radius
Let the center of the circumcircle be \( (h, k) \) and the radius be \( r \). Since the circumcircle touches the x-axis, we have:
\[ k = r \]

### Step 3: Find the Coordinates of the Center
The circumcenter of triangle \( ABC \) can be found using the perpendicular bisectors of the sides. However, we can also use the fact that the circumradius \( R \) can be expressed in terms of the triangle's vertices.

### Step 4: Use the Distance Formula
The distances from the circumcenter to each vertex must equal the circumradius \( r \). We can set up the following equations based on the vertices:

1. Distance from \( (h, k) \) to \( A(0, 3) \):
\[ \sqrt{(h - 0)^2 + (k - 3)^2} = r \]
Squaring both sides:
\[ h^2 + (k - 3)^2 = r^2 \]

2. Distance from \( (h, k) \) to \( B(0, 12) \):
\[ \sqrt{(h - 0)^2 + (k - 12)^2} = r \]
Squaring both sides:
\[ h^2 + (k - 12)^2 = r^2 \]

3. Distance from \( (h, k) \) to \( C(x, 0) \):
\[ \sqrt{(h - x)^2 + (k - 0)^2} = r \]
Squaring both sides:
\[ (h - x)^2 + k^2 = r^2 \]

### Step 5: Set Up the Equations
From the first two equations, we can set them equal to each other:
\[ h^2 + (k - 3)^2 = h^2 + (k - 12)^2 \]
This simplifies to:
\[ (k - 3)^2 = (k - 12)^2 \]
Expanding both sides:
\[ k^2 - 6k + 9 = k^2 - 24k + 144 \]
Simplifying gives:
\[ 18k = 135 \implies k = 7.5 \]
Thus, the radius \( r = 7.5 \).

### Step 6: Find the Value of \( x \)
Using \( k = 7.5 \) in the first equation:
\[ h^2 + (7.5 - 3)^2 = (7.5)^2 \]
This gives:
\[ h^2 + 20.25 = 56.25 \implies h^2 = 36 \implies h = 6 \text{ or } -6 \]

### Step 7: Calculate the Area of Triangle \( ABC \)
The coordinates of the vertices are:
- \( A(0, 3) \)
- \( B(0, 12) \)
- \( C(6, 0) \)

Using the formula for the area of a triangle given its vertices:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0(12 - 0) + 0(0 - 3) + 6(3 - 12) \right| \]
This simplifies to:
\[ \text{Area} = \frac{1}{2} \left| 6(-9) \right| = \frac{1}{2} \times 54 = 27 \text{ sq. units} \]

### Final Answer
The area of triangle \( ABC \) is \( 27 \) square units.