A circle of radius r, `rne0` touches the parabola `y^(2)+12x=0` at the vertex of the parabola. The centre of the circle lies to the left of the vertex and the circle lies entirely witin the parabola. Then the interval(s) in which r lies can be

Options

A. (1, 7)
B. (0, 6)
C. `(1, 13/2)`
D. (2, 8)

Answer

To solve the problem, we need to find the interval in which the radius \( r \) of a circle, which touches the parabola \( y^2 + 12x = 0 \) at its vertex, can lie. The center of the circle is to the left of the vertex and the circle lies entirely within the parabola.

### Step-by-step Solution:

1. **Identify the Parabola**:
The given parabola is \( y^2 + 12x = 0 \) or \( y^2 = -12x \). This is a left-opening parabola with its vertex at the origin (0,0).

**Hint**: Remember that the vertex of the parabola is the point where it changes direction.

2. **Circle's Center and Radius**:
The circle has a radius \( r \) and touches the parabola at the vertex (0,0). Since the center of the circle lies to the left of the vertex, we can denote the center of the circle as \( (-r, 0) \).

**Hint**: The center of the circle is at a distance \( r \) from the vertex along the x-axis.

3. **Equation of the Circle**:
The equation of the circle with center at \( (-r, 0) \) and radius \( r \) is given by:
\[ (x + r)^2 + y^2 = r^2 \]

**Hint**: Substitute the coordinates of the center and the radius into the standard equation of a circle.

4. **Finding Intersection Points**:
To find the intersection points of the circle and the parabola, substitute \( y^2 = -12x \) into the circle's equation:
\[ (x + r)^2 + (-12x) = r^2 \]
Expanding this gives:
\[ (x + r)^2 - 12x = r^2 \]
\[ x^2 + 2rx + r^2 - 12x = r^2 \]
Simplifying, we have:
\[ x^2 + (2r - 12)x = 0 \]

**Hint**: This is a quadratic equation in \( x \). Factor it to find the roots.

5. **Finding Roots**:
Factor out \( x \):
\[ x(x + 2r - 12) = 0 \]
The roots are \( x = 0 \) and \( x = 12 - 2r \).

**Hint**: The roots represent the points where the circle intersects the parabola.

6. **Condition for No Other Intersection**:
For the circle to lie entirely within the parabola, the second root \( 12 - 2r \) must be less than or equal to zero (it cannot intersect again):
\[ 12 - 2r \leq 0 \]
Solving this gives:
\[ 2r \geq 12 \implies r \geq 6 \]

**Hint**: This condition ensures that the circle does not intersect the parabola again.

7. **Radius Constraint**:
Since \( r \) is a radius, it must also be positive:
\[ r > 0 \]

**Hint**: The radius cannot be negative or zero.

8. **Final Interval**:
Combining the two inequalities, we have:
\[ 0 < r < 6 \]

Therefore, the interval in which \( r \) lies is \( (0, 6) \).

### Conclusion:
The radius \( r \) of the circle lies in the interval \( (0, 6) \).