If two perpendicular tangents can be drawn from the origin to the circle `x^2-6x+y^2-2p y+17=0` , then the value of `|p|` is___

Explanation

To solve the problem, we need to find the value of |p| given that two perpendicular tangents can be drawn from the origin to the circle defined by the equation \( x^2 - 6x + y^2 - 2py + 17 = 0 \).

### Step-by-Step Solution:

1. **Rewrite the Circle Equation**:
The given equation of the circle is:
\[ x^2 - 6x + y^2 - 2py + 17 = 0 \]
We will complete the square for both \(x\) and \(y\).

2. **Complete the Square for \(x\)**:
For \(x^2 - 6x\):
\[ x^2 - 6x = (x - 3)^2 - 9 \]

3. **Complete the Square for \(y\)**:
For \(y^2 - 2py\):
\[ y^2 - 2py = (y - p)^2 - p^2 \]

4. **Substitute Back into the Circle Equation**:
Substitute the completed squares back into the equation:
\[ (x - 3)^2 - 9 + (y - p)^2 - p^2 + 17 = 0 \]
Simplifying this gives:
\[ (x - 3)^2 + (y - p)^2 - (9 + p^2 - 17) = 0 \]
\[ (x - 3)^2 + (y - p)^2 = p^2 - 8 \]

5. **Identify the Center and Radius**:
The center of the circle is \((3, p)\) and the radius \(r\) is \(\sqrt{p^2 - 8}\).

6. **Equation of the Director Circle**:
The equation of the director circle is given by:
\[ x^2 + y^2 = 2r^2 \]
Substituting for \(r\):
\[ x^2 + y^2 = 2(p^2 - 8) \]

7. **Condition for Perpendicular Tangents**:
Since the tangents from the origin (0,0) to the circle are perpendicular, the point (0,0) must satisfy the director circle equation:
\[ 0^2 + 0^2 = 2(p^2 - 8) \]
This simplifies to:
\[ 0 = 2(p^2 - 8) \]
Thus:
\[ p^2 - 8 = 0 \implies p^2 = 8 \]

8. **Finding |p|**:
Taking the square root gives:
\[ p = \pm \sqrt{8} = \pm 2\sqrt{2} \]
Therefore, the value of \(|p|\) is:
\[ |p| = 2\sqrt{2} \]

### Final Answer:
\[ |p| = 2\sqrt{2} \]