For the circle `x^(2) + y^(2) = 25,` the point (-2.5, 3.5) lies :

Options

A. Inside circle
B. Outside circle
C. On the circle
D. none of these.

Answer

To determine the position of the point (-2.5, 3.5) relative to the circle defined by the equation \( x^2 + y^2 = 25 \), we will follow these steps:

### Step 1: Identify the center and radius of the circle
The equation of the circle is given as:
\[ x^2 + y^2 = 25 \]
From this equation, we can identify that:
- The center of the circle \( C \) is at \( (0, 0) \).
- The radius \( R \) of the circle is \( \sqrt{25} = 5 \).

**Hint:** The standard form of a circle's equation is \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and \( r \) is the radius.

### Step 2: Calculate the distance from the point to the center of the circle
We need to calculate the distance \( PC \) from the point \( P(-2.5, 3.5) \) to the center \( C(0, 0) \) using the distance formula:
\[ PC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
Substituting the coordinates of \( P \) and \( C \):
\[ PC = \sqrt{(-2.5 - 0)^2 + (3.5 - 0)^2} \]
\[ PC = \sqrt{(-2.5)^2 + (3.5)^2} \]

**Hint:** The distance formula is derived from the Pythagorean theorem.

### Step 3: Simplify the distance calculation
Calculating the squares:
\[ PC = \sqrt{6.25 + 12.25} \]
\[ PC = \sqrt{18.5} \]

**Hint:** Remember that \( a^2 + b^2 = c^2 \) is used here to find the hypotenuse in a right triangle formed by the coordinates.

### Step 4: Compare the distance with the radius
Now, we need to compare \( PC \) with the radius \( R \):
- We found \( R = 5 \).
- We need to evaluate \( \sqrt{18.5} \).

Calculating \( \sqrt{18.5} \):
\[ \sqrt{18.5} \approx 4.3 \]

Now, we compare:
\[ PC \approx 4.3 < R = 5 \]

**Hint:** If the distance from the point to the center is less than the radius, the point lies inside the circle.

### Conclusion
Since \( PC < R \), the point (-2.5, 3.5) lies **inside the circle**.

**Final Answer:** The point (-2.5, 3.5) lies inside the circle.