The equation of an ex - circle of a triangle formed by the common tangents to the circle `x^(2)+y^(2)=4 and x^(2)+y^(2)-6x+8=0` is

Options

A. `x^(2)+y^(2)=4`
B. `x^(2)+y^(2)-6x+8=0`
C. `x^(2)+y^(2)-6x+9=0`
D. `x^(2)+y^(2)=1`

Answer

To find the equation of the ex-circle of the triangle formed by the common tangents to the circles given by the equations \(x^2 + y^2 = 4\) and \(x^2 + y^2 - 6x + 8 = 0\), we can follow these steps:

### Step 1: Identify the centers and radii of the circles

1. **First Circle**: The equation \(x^2 + y^2 = 4\) can be rewritten in standard form. Here, the center is at \((0, 0)\) and the radius is \(r_1 = 2\) (since \(4 = 2^2\)).

2. **Second Circle**: The equation \(x^2 + y^2 - 6x + 8 = 0\) can be rearranged:
\[ x^2 - 6x + y^2 + 8 = 0 \implies (x^2 - 6x + 9) + y^2 = 1 \implies (x - 3)^2 + y^2 = 1 \]
This shows that the center is at \((3, 0)\) and the radius is \(r_2 = 1\) (since \(1 = 1^2\)).

### Step 2: Determine the distance between the centers

The distance \(d\) between the centers of the two circles \((0, 0)\) and \((3, 0)\) is:
\[ d = \sqrt{(3 - 0)^2 + (0 - 0)^2} = \sqrt{9} = 3 \]

### Step 3: Find the radius of the ex-circle

The radius \(R\) of the ex-circle can be calculated using the formula:
\[ R = \frac{d + r_1 + r_2}{2} \]
Substituting the known values:
\[ R = \frac{3 + 2 + 1}{2} = \frac{6}{2} = 3 \]

### Step 4: Write the equation of the ex-circle

The ex-circle is centered at the point that is external to the triangle formed by the tangents to the circles. Since the centers of the circles are \((0, 0)\) and \((3, 0)\), and the ex-circle is external, its center will be at the midpoint of the line connecting the two centers plus the radius of the larger circle:
\[ \text{Center of ex-circle} = (3 + 3, 0) = (6, 0) \]

The equation of the ex-circle with center \((6, 0)\) and radius \(3\) is given by:
\[ (x - 6)^2 + (y - 0)^2 = 3^2 \implies (x - 6)^2 + y^2 = 9 \]

### Final Answer

Thus, the equation of the ex-circle is:
\[ (x - 6)^2 + y^2 = 9 \]