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The problem you’ve shared involves determining when a line is tangent to a circle, using the concept of the discriminant. Let's break it down:

### Problem Summary:
- You are given a **line equation**: 
  \[
  2x + y - 5 = 0
  \]
  and a **circle equation**: 
  \[
  (x - 3)^2 + (y - p)^2 = 5
  \]
- The problem is to:
  1. **Find the values of \( p \)** such that the line is tangent to the circle.
  2. **Write down the coordinates of the center of the circle** in each case.

### Step 1: Substituting the line equation into the circle equation
We will substitute \( y = -2x + 5 \) from the line into the circle equation to form a quadratic in \( x \):

The circle equation is:
\[
(x - 3)^2 + (y - p)^2 = 5
\]
Substitute \( y = -2x + 5 \):
\[
(x - 3)^2 + (-2x + 5 - p)^2 = 5
\]

### Step 2: Expanding and simplifying
1. Expand \( (x - 3)^2 \) and \( (-2x + 5 - p)^2 \):
   \[
   (x - 3)^2 = x^2 - 6x + 9
   \]
   \[
   (-2x + 5 - p)^2 = (4x^2 - 20x + 25 - 10p + p^2)
   \]
   
2. Plug these back into the equation:
   \[
   x^2 - 6x + 9 + 4x^2 - 20x + 25 - 10p + p^2 = 5
   \]
   Simplify:
   \[
   5x^2 - 26x + (34 - 10p + p^2) = 5
   \]
   Subtract 5 from both sides:
   \[
   5x^2 - 26x + (29 - 10p + p^2) = 0
   \]

### Step 3: Use the discriminant to find the condition for tangency
For the line to be tangent to the circle, the quadratic must have exactly **one solution**, which means the discriminant must be zero:
\[
\Delta = b^2 - 4ac = 0
\]
Here, \( a = 5 \), \( b = -26 \), and \( c = 29 - 10p + p^2 \).

1. Compute the discriminant:
   \[
   \Delta = (-26)^2 - 4(5)(29 - 10p + p^2)
   \]
   \[
   \Delta = 676 - 20(29 - 10p + p^2)
   \]
   \[
   \Delta = 676 - 580 + 200p - 20p^2
   \]
   \[
   \Delta = 96 + 200p - 20p^2
   \]
   Set \( \Delta = 0 \) for tangency:
   \[
   96 + 200p - 20p^2 = 0
   \]
   Divide by 4 to simplify:
   \[
   24 + 50p - 5p^2 = 0
   \]
   Multiply by -1 to rearrange:
   \[
   5p^2 - 50p - 24 = 0
   \]

### Step 4: Solve the quadratic equation for \( p \)
Solve using the quadratic formula:
\[
p = \frac{-(-50) \pm \sqrt{(-50)^2 - 4(5)(-24)}}{2(5)}
\]
\[
p = \frac{50 \pm \sqrt{2500 + 480}}{10}
\]
\[
p = \frac{50 \pm \sqrt{2980}}{10}
\]
\[
p = \frac{50 \pm 54.6}{10}
\]
Thus, the two possible values of \( p \) are:
\[
p_1 = \frac{50 + 54.6}{10} = 10.46, \quad p_2 = \frac{50 - 54.6}{10} = -0.46
\]

### Step 5: Coordinates of the center of the circle
For each value of \( p \), the coordinates of the center of the circle are:
- When \( p_1 = 10.46 \), the center is \( (3, 10.46) \)
- When \( p_2 = -0.46 \), the center is \( (3, -0.46) \)

---

Would you like a deeper explanation of any step or any further details?

### Related Questions:
1. How do you derive the equation of a line tangent to a circle in general form?
2. How does the discriminant help in determining the nature of solutions to a quadratic equation?
3. Can you solve for other values of \( p \) if the line intersects the circle at two points?
4. How does changing the center of the circle affect the condition for tangency?
5. What happens to the quadratic discriminant when a line does not intersect the circle?

#### Tip:
Always simplify your discriminant equation as much as possible before solving; it often reduces calculation errors.

The line with equation 2x + y - 5 = 0 is a tangent to the circle with equation (x - 3)^2 + (y - p)^2 = 5. Find the two possible values for p. Write down the coordinates of the center of the circle in each case.

Learn how to find the values of p such that the line 2x + y - 5 = 0 is tangent to the circle (x - 3)^2 + (y - p)^2 = 5. This solution uses algebraic techniques involving the discriminant of a quadratic equation, suitable for students in Grades 9-11.

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