If the parabola `x^(2)=4y` and the cilcle `x^(2)+(y-7)^(2)=r^(2)` have maxiumum number of common chords , then least value of r is `(sqrt6=2.45)`

Answer

To find the least value of \( r \) such that the parabola \( x^2 = 4y \) and the circle \( x^2 + (y - 7)^2 = r^2 \) have the maximum number of common chords, we can follow these steps:

### Step 1: Understand the Equations
The parabola is given by the equation \( x^2 = 4y \), which opens upwards. The circle is defined by the equation \( x^2 + (y - 7)^2 = r^2 \), which is centered at \( (0, 7) \) with radius \( r \).

**Hint:** Identify the characteristics of both the parabola and the circle, such as their shapes and positions on the coordinate plane.

### Step 2: Substitute \( y \) from the Parabola into the Circle's Equation
From the parabola, we can express \( y \) in terms of \( x \):
\[ y = \frac{x^2}{4} \]
Now, substitute this into the circle's equation:
\[ x^2 + \left(\frac{x^2}{4} - 7\right)^2 = r^2 \]

**Hint:** This substitution will help us find the points of intersection between the two curves.

### Step 3: Expand the Circle's Equation
Expanding the equation gives:
\[ x^2 + \left(\frac{x^2}{4} - 7\right)^2 = r^2 \]
Expanding \( \left(\frac{x^2}{4} - 7\right)^2 \):
\[ \left(\frac{x^2}{4} - 7\right)^2 = \frac{x^4}{16} - \frac{7x^2}{2} + 49 \]
Thus, the equation becomes:
\[ x^2 + \frac{x^4}{16} - \frac{7x^2}{2} + 49 = r^2 \]
Rearranging gives:
\[ \frac{x^4}{16} + \left(1 - \frac{7}{2}\right)x^2 + 49 - r^2 = 0 \]
This simplifies to:
\[ \frac{x^4}{16} - \frac{5}{2}x^2 + (49 - r^2) = 0 \]

**Hint:** This is a quadratic equation in \( x^2 \).

### Step 4: Set the Discriminant Greater Than or Equal to Zero
For the quadratic equation \( ax^2 + bx + c = 0 \) to have real roots, the discriminant must be non-negative:
\[ D = b^2 - 4ac \geq 0 \]
Here, \( a = \frac{1}{16} \), \( b = -\frac{5}{2} \), and \( c = 49 - r^2 \). The discriminant is:
\[ \left(-\frac{5}{2}\right)^2 - 4 \cdot \frac{1}{16} \cdot (49 - r^2) \geq 0 \]
Calculating gives:
\[ \frac{25}{4} - \frac{1}{4}(49 - r^2) \geq 0 \]
Multiplying through by 4 to eliminate the fraction:
\[ 25 - (49 - r^2) \geq 0 \]
This simplifies to:
\[ r^2 - 24 \geq 0 \]

**Hint:** This condition ensures that the circle intersects the parabola in real points.

### Step 5: Solve for \( r \)
From \( r^2 - 24 \geq 0 \):
\[ r^2 \geq 24 \]
Taking the square root gives:
\[ r \geq 2\sqrt{6} \]

**Hint:** Remember that \( r \) must be positive, so we only consider the positive root.

### Step 6: Conclusion
The least value of \( r \) for which the parabola and the circle have the maximum number of common chords is:
\[ r = 2\sqrt{6} \]
Numerically, \( 2\sqrt{6} \approx 4.9 \).

**Final Answer:** The least value of \( r \) is \( 2\sqrt{6} \) or approximately \( 4.9 \).