C is the centre of the circle with centre `(0,1)` and radius unity. `y=ax^2` is a parabola. The set of the values of `'a'` for which they meet at a point other than the origin, is

Options

A. `agt0`
B. `ain(0,1//2)`
C. (1/4, 1/2)
D. `(1//2,oo)`

Explanation

To solve the problem, we need to find the values of \( a \) for which the parabola \( y = ax^2 \) intersects the circle centered at \( (0, 1) \) with a radius of 1 at a point other than the origin.

### Step-by-Step Solution:

1. **Write the equation of the circle**:
The equation of a circle with center \( (0, 1) \) and radius 1 is given by:
\[ (x - 0)^2 + (y - 1)^2 = 1 \]
Simplifying this, we get:
\[ x^2 + (y - 1)^2 = 1 \]

2. **Expand the equation of the circle**:
Expanding the equation:
\[ x^2 + (y^2 - 2y + 1) = 1 \]
This simplifies to:
\[ x^2 + y^2 - 2y = 0 \]

3. **Substitute the parabola equation into the circle equation**:
We know that \( y = ax^2 \). Substituting this into the circle's equation:
\[ x^2 + (ax^2)^2 - 2(ax^2) = 0 \]
This becomes:
\[ x^2 + a^2x^4 - 2ax^2 = 0 \]
Rearranging gives:
\[ a^2x^4 + (1 - 2a)x^2 = 0 \]

4. **Factor out \( x^2 \)**:
Factoring out \( x^2 \):
\[ x^2(a^2x^2 + (1 - 2a)) = 0 \]
This gives us two solutions:
- \( x^2 = 0 \) (which corresponds to the origin)
- \( a^2x^2 + (1 - 2a) = 0 \) (the other intersection point)

5. **Find conditions for the second intersection point**:
For the second intersection point to exist, we need:
\[ 1 - 2a > 0 \]
Solving this inequality:
\[ 1 > 2a \implies a < \frac{1}{2} \]

6. **Check for the positivity of \( a^2 \)**:
Since \( a^2 \) is always non-negative, we do not have any restrictions from this part. Thus, we only need to consider the condition \( a < \frac{1}{2} \).

7. **Conclusion**:
Therefore, the set of values of \( a \) for which the parabola intersects the circle at a point other than the origin is:
\[ a < \frac{1}{2} \]

### Final Answer:
The set of values of \( a \) is \( (-\infty, \frac{1}{2}) \).