The point (p,0) lies on a circle in the xy-plane. The points (2,3.5) and (-2,0.5) are the endpoints of a diameter of the circle . If `p gt 0`, what is the value of p?

Answer

To solve the problem, we need to find the value of \( p \) such that the point \( (p, 0) \) lies on a circle defined by the endpoints of its diameter, which are \( (2, 3.5) \) and \( (-2, 0.5) \).

### Step 1: Find the equation of the circle using the endpoints of the diameter.

The general equation of a circle given two endpoints of a diameter \( (x_1, y_1) \) and \( (x_2, y_2) \) can be expressed as:

\[ (x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0 \]

Here, we have:
- \( x_1 = 2 \)
- \( y_1 = 3.5 \)
- \( x_2 = -2 \)
- \( y_2 = 0.5 \)

Substituting these values into the equation:

\[ (x - 2)(x + 2) + (y - 3.5)(y - 0.5) = 0 \]

### Step 2: Expand the equation.

First, expand \( (x - 2)(x + 2) \):

\[ (x - 2)(x + 2) = x^2 - 4 \]

Next, expand \( (y - 3.5)(y - 0.5) \):

\[ (y - 3.5)(y - 0.5) = y^2 - 4y + 1.75 \]

Now combine these results:

\[ x^2 - 4 + y^2 - 4y + 1.75 = 0 \]

### Step 3: Rearrange the equation.

Combine like terms:

\[ x^2 + y^2 - 4y - 2.25 = 0 \]

### Step 4: Substitute the point \( (p, 0) \) into the equation.

Since the point \( (p, 0) \) lies on the circle, we substitute \( x = p \) and \( y = 0 \):

\[ p^2 + 0^2 - 4(0) - 2.25 = 0 \]

This simplifies to:

\[ p^2 - 2.25 = 0 \]

### Step 5: Solve for \( p \).

Rearranging gives:

\[ p^2 = 2.25 \]

Taking the square root of both sides:

\[ p = \sqrt{2.25} = 1.5 \]

Since we are given that \( p > 0 \), we conclude:

\[ p = 1.5 \]

### Final Answer:

The value of \( p \) is \( 1.5 \).

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