STATEMENT-1 : Equation of circle which touches the circle `x^(2) + y^(2) - 6x +6y + 17 = 0` externally and to which the lines `x^(2) - 3xy - 3x + 9y = 0` are normal is `x^(2) + y^(2) - 6x - 2y +1 = 0`.
STATEMENT-2 : Equation of circle which touches the circle `x^(2) + y^(2) -6x + 6y + 17 = 0` internally and to which the line `x^(2) - 3xy - 3x + 9y = 0` are normal is `x^(2) + y^(2) -6x - 2y -15 = 0`.
STATMENT-3 : Equation of circle which is orthogonal to circle `x^(2) + y^(2) -6x + 6y + 17 = 0` and have normals along `x^(2) -3xy -3x + 9y =0` is `x^(2) + y^(2) - 6x -2 y-5 = 0`.

Options

A. T T F
B. T F F
C. F F F
D. T T T

Answer

To solve the problem, we need to analyze each statement regarding the circles and their properties. Let's break down the solution step by step.

### Step 1: Analyze Statement 1
We have the first circle given by the equation:
\[ x^2 + y^2 - 6x + 6y + 17 = 0 \]

**Finding the center and radius of the first circle:**
1. Rewrite the equation in standard form:
\[ (x^2 - 6x) + (y^2 + 6y) + 17 = 0 \]
Completing the square:
\[ (x - 3)^2 - 9 + (y + 3)^2 - 9 + 17 = 0 \]
\[ (x - 3)^2 + (y + 3)^2 - 1 = 0 \]
Thus, the center is \( C_1(3, -3) \) and the radius \( r_1 = 1 \).

**Finding the equation of the second circle:**
2. The second circle touches the first circle externally. Let the center of the second circle be \( C_2(a, b) \) and its radius be \( r_2 \). The condition for external tangency is:
\[ |C_1C_2| = r_1 + r_2 \]
The distance \( |C_1C_2| \) is given by:
\[ \sqrt{(a - 3)^2 + (b + 3)^2} \]

3. The lines given by:
\[ x^2 - 3xy - 3x + 9y = 0 \]
can be factored to find the slopes of the normals. Solving gives us the points where the normals intersect.

4. The equation of the second circle is given as:
\[ x^2 + y^2 - 6x - 2y + 1 = 0 \]
This can be rewritten to find its center and radius:
\[ (x - 3)^2 + (y - 1)^2 = 1 \]
Thus, the center is \( C_2(3, 1) \) and radius \( r_2 = 1 \).

5. Check the distance condition:
\[ |C_1C_2| = \sqrt{(3 - 3)^2 + (1 + 3)^2} = \sqrt{16} = 4 \]
The sum of the radii:
\[ r_1 + r_2 = 1 + 1 = 2 \]
Since \( 4 \neq 2 \), Statement 1 is **incorrect**.

### Step 2: Analyze Statement 2
For the second statement, we need to find a circle that touches the first circle internally.

1. The condition for internal tangency is:
\[ |C_1C_2| = |r_1 - r_2| \]
Using the same centers and radius as before, we find:
\[ |C_1C_2| = 4 \]
Setting up the equation:
\[ 4 = |1 - r_2| \]
This gives two cases:
- \( 1 - r_2 = 4 \) → \( r_2 = -3 \) (not possible)
- \( r_2 - 1 = 4 \) → \( r_2 = 5 \)

2. The equation of the circle is:
\[ (x - 3)^2 + (y - 1)^2 = 25 \]
This simplifies to:
\[ x^2 + y^2 - 6x - 2y - 15 = 0 \]
This matches the given equation for Statement 2, so it is **correct**.

### Step 3: Analyze Statement 3
For the third statement, we need to find a circle orthogonal to the first circle.

1. The condition for orthogonality is:
\[ |C_1C_2|^2 = r_1^2 + r_2^2 \]
We already know:
\[ |C_1C_2| = 4 \quad \text{and} \quad r_1 = 1 \]
Thus:
\[ 16 = 1 + r_2^2 \quad \Rightarrow \quad r_2^2 = 15 \quad \Rightarrow \quad r_2 = \sqrt{15} \]

2. The equation of the circle becomes:
\[ (x - 3)^2 + (y - 1)^2 = 15 \]
This simplifies to:
\[ x^2 + y^2 - 6x - 2y - 5 = 0 \]
This matches the given equation for Statement 3, so it is also **correct**.

### Conclusion
- Statement 1: **Incorrect**
- Statement 2: **Correct**
- Statement 3: **Correct**