There is a point inside a circle. What is the probability that this point is close to the circumference than to the centre ?

Options

A. `3//4`
B. `1//2`
C. `1//4`
D. `1//3`

Explanation

To solve the problem of finding the probability that a point inside a circle is closer to the circumference than to the center, we can follow these steps:

### Step 1: Understand the Problem
We need to find the area where points are closer to the circumference than to the center of the circle.

### Step 2: Define the Circle
Let the radius of the circle be \( r \). The center of the circle is point \( O \) and the circumference is at a distance \( r \) from \( O \).

### Step 3: Determine the Distance Criteria
A point \( P \) inside the circle is closer to the circumference than to the center if the distance from \( P \) to the circumference is less than the distance from \( P \) to the center.

### Step 4: Set Up the Inequality
If the distance from point \( P \) to the center \( O \) is \( d \), then the distance from \( P \) to the circumference is \( r - d \). We need to find when:
\[ r - d < d \]
This simplifies to:
\[ r < 2d \quad \Rightarrow \quad d > \frac{r}{2} \]

### Step 5: Identify the Relevant Area
The points that satisfy \( d > \frac{r}{2} \) lie outside a smaller circle with radius \( \frac{r}{2} \) centered at \( O \).

### Step 6: Calculate Areas
1. **Area of the larger circle** (radius \( r \)):
\[ A_{\text{large}} = \pi r^2 \]

2. **Area of the smaller circle** (radius \( \frac{r}{2} \)):
\[ A_{\text{small}} = \pi \left(\frac{r}{2}\right)^2 = \pi \frac{r^2}{4} \]

3. **Area of the annular region** (the area between the two circles):
\[ A_{\text{annulus}} = A_{\text{large}} - A_{\text{small}} = \pi r^2 - \pi \frac{r^2}{4} = \pi r^2 \left(1 - \frac{1}{4}\right) = \pi r^2 \frac{3}{4} \]

### Step 7: Calculate the Probability
The probability that a randomly chosen point inside the circle is closer to the circumference than to the center is given by the ratio of the area of the annular region to the area of the larger circle:
\[ P = \frac{A_{\text{annulus}}}{A_{\text{large}}} = \frac{\frac{3\pi r^2}{4}}{\pi r^2} = \frac{3}{4} \]

### Conclusion
Thus, the required probability is:
\[ \boxed{\frac{3}{4}} \]

---