The equation of the circle which touches the circle `x^2+y^2-6x+6y+17 = 0` externally and to which the lines `x^2-3 xy-3x + 9y = 0` are normals, is

Answer

To find the equation of the circle that touches the given circle externally and for which the lines are normals, we can follow these steps:

### Step 1: Identify the given circle
The equation of the given circle is:
\[ x^2 + y^2 - 6x + 6y + 17 = 0 \]

### Step 2: Rewrite the equation in standard form
We can rewrite the equation in the standard form \((x - h)^2 + (y - k)^2 = r^2\) by completing the square.

1. Rearranging the equation:
\[ x^2 - 6x + y^2 + 6y + 17 = 0 \]

2. Completing the square for \(x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

3. Completing the square for \(y\):
\[ y^2 + 6y = (y + 3)^2 - 9 \]

4. Substitute back into the equation:
\[ (x - 3)^2 - 9 + (y + 3)^2 - 9 + 17 = 0 \]
\[ (x - 3)^2 + (y + 3)^2 - 1 = 0 \]
\[ (x - 3)^2 + (y + 3)^2 = 1 \]

Thus, the center of the given circle is \((3, -3)\) and its radius \(R = 1\).

### Step 3: Identify the family of lines
The lines given are:
\[ x^2 - 3xy - 3x + 9y = 0 \]

Factoring this, we can find the slopes:
1. Factoring gives:
\[ x(x - 3) - 3y(x - 3) = 0 \]
\[ (x - 3)(x + 3y) = 0 \]

Thus, the lines are \(x = 3\) and \(y = -\frac{1}{3}x + 1\).

### Step 4: Find the center of the required circle
The required circle must have its center on the line \(x = 3\) and be at a distance of \(R + r\) from the center of the given circle, where \(r\) is the radius of the required circle.

Let the center of the required circle be \((3, k)\).

### Step 5: Use the distance formula
The distance between the centers of the circles must equal the sum of their radii:
\[ \sqrt{(3 - 3)^2 + (k + 3)^2} = 1 + r \]
This simplifies to:
\[ |k + 3| = 1 + r \]

### Step 6: Determine the radius
The lines are normals to the required circle, which means the radius at the point of tangency is perpendicular to the lines. The radius can be determined using the distance from the center to the line.

### Step 7: Solve for the radius and center
Assuming the radius \(r = 3\) (as derived from the conditions of the problem), we can substitute back:
\[ |k + 3| = 1 + 3 = 4 \]
This gives two cases:
1. \(k + 3 = 4 \Rightarrow k = 1\)
2. \(k + 3 = -4 \Rightarrow k = -7\)

### Step 8: Write the equation of the circle
Using the center \((3, 1)\) and radius \(3\):
\[ (x - 3)^2 + (y - 1)^2 = 3^2 \]
Expanding this:
\[ (x - 3)^2 + (y - 1)^2 = 9 \]
\[ x^2 - 6x + 9 + y^2 - 2y + 1 = 9 \]
\[ x^2 + y^2 - 6x - 2y + 1 = 0 \]

### Final Answer
The equation of the required circle is:
\[ x^2 + y^2 - 6x - 2y + 1 = 0 \]