1. Understanding the Problem and Key Concepts

This problem challenges us to find a line that simultaneously acts as a common tangent to two given parabolas, $y^2 = 4x$ and $x^2 = 4y$. Once we determine this unique common tangent line, we then need to use the condition that this very line also touches a given circle, $x^2 + y^2 = c^2$, to find the value of $c$. This requires a strong understanding of tangent properties for conic sections and the condition for a line to be tangent to a circle.

To solve this problem effectively, we will utilize the following fundamental concepts and formulas from coordinate geometry:

• Equation of a Tangent to a Parabola $y^2 = 4ax$ (Slope Form): If a line with slope $m$ is tangent to the parabola $y^2 = 4ax$, its equation is given by: $y = mx + \frac{a}{m}$ Rationale: This formula allows us to express any tangent line to this type of parabola solely in terms of its slope $m$ and the parabola's focal parameter $a$.

• Equation of a Tangent to a Parabola $x^2 = 4ay$ (Slope Form): Similarly, if a line with slope $m$ is tangent to the parabola $x^2 = 4ay$, its equation is given by: $y = mx - am^2$ Rationale: This is the corresponding slope-form tangent equation for parabolas opening upwards or downwards. It's crucial to distinguish this from the previous form.

• Condition for Tangency of a Line to a Circle: A straight line $Ax + By + C = 0$ is tangent to a circle with center $(x_0, y_0)$ and radius $r$ if and only if the perpendicular distance from the center of the circle to the line is equal to the radius $r$. The distance $d$ is calculated using the formula: $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$ For a circle centered at the origin $(0,0)$ with radius $r$, this condition simplifies to: $r = \frac{|C|}{\sqrt{A^2 + B^2}}$ Rationale: This geometric condition is the definitive test for tangency. If the distance from the center to the line is less than the radius, the line intersects the circle; if it's greater, it doesn't intersect. Only when it equals the radius is it a tangent.

2. Step-by-Step Solution

Our strategy will involve finding the general tangent equations for both parabolas, equating them to find the common slope, determining the common tangent line, and then applying the circle tangency condition.

Step 1: Identify Parameters and Write General Tangent Equations for Both Parabolas

We begin by identifying the parameter 'a' for each parabola and then writing down the general equation of a tangent line with slope $m$ for each. This is a standard approach when seeking a common tangent, as the common tangent will necessarily have a single, common slope $m$.

• For the parabola $y^2 = 4x$: This parabola is in the standard form $y^2 = 4ax$. Comparing $y^2 = 4x$ with $y^2 = 4ax$, we can see that: $4a = 4 \implies a = 1$ Now, using the formula for a tangent with slope $m$ to $y^2 = 4ax$, we substitute $a=1$: $y = mx + \frac{1}{m} \quad \text{(Equation 1)}$ Explanation: This equation represents all possible tangent lines with slope $m$ to the parabola $y^2=4x$. We assume our common tangent has some slope $m$.

• For the parabola $x^2 = 4y$: This parabola is in the standard form $x^2 = 4ay$. Comparing $x^2 = 4y$ with $x^2 = 4ay$, we again find: $4a = 4 \implies a = 1$ Next, we use the formula for a tangent with slope $m$ to $x^2 = 4ay$, substituting $a=1$: $y = mx - (1)m^2$ $y = mx - m^2 \quad \text{(Equation 2)}$ Explanation: Similarly, this equation represents all possible tangent lines with slope $m$ to the parabola $x^2=4y$. Since we are searching for a common tangent, it must have the same slope $m$ for both parabolas.

Step 2: Find the Slope of the Common Tangent

For a line to be a common tangent to both parabolas, it must be the same line. Since we have already assumed that the slopes ($m$) of the tangent lines are equal, for them to represent the same line, their y-intercepts must also be identical.

Equating the y-intercepts from Equation 1 and Equation 2: $\frac{1}{m} = -m^2$ Explanation: If two linear equations $y = mx + c_1$ and $y = mx + c_2$ represent the same line, and they already share the same slope $m$, then their y-intercepts ($c_1$ and $c_2$) must necessarily be identical. This crucial step allows us to solve for the specific slope $m$ that satisfies the condition of being a common tangent.

Now, we solve this equation for $m$: $1 = -m^3$ $m^3 = -1$ Taking the cube root of both sides, we find the only real solution for $m$: $m = -1$ Explanation: In coordinate geometry, a tangent line must have a real slope. The cube root of $-1$ is $-1$, which is a real number, confirming that a common tangent exists.

Step 3: Determine the Equation of the Common Tangent

Now that we have found the common slope $m = -1$, we can substitute this value into either Equation 1 or Equation 2 to obtain the specific equation of the common tangent line. Let's use Equation 1: $y = (-1)x + \frac{1}{(-1)}$ $y = -x - 1$ To prepare for applying the tangency condition with the circle, we rearrange this equation into the standard linear form $Ax + By + C = 0$: $x + y + 1 = 0 \quad \text{(Equation of Common Tangent)}$ Explanation: Substituting the specific common slope $m=-1$ into one of the general tangent equations yields the unique equation of the line that touches both parabolas simultaneously. This is the line we're interested in.

Step 4: Apply the Tangency Condition for the Circle

The problem states that this common tangent line, $x + y + 1 = 0$, also touches the circle $x^2 + y^2 = c^2$. Let's identify the properties of this circle:

• The circle $x^2 + y^2 = c^2$ is centered at the origin, so $(x_0, y_0) = (0,0)$.
• The radius of the circle is $r = c$. (Note: radius must be positive, so $c>0$).

For the line $x + y + 1 = 0$ to be tangent to this circle, the perpendicular distance from the center of the circle $(0,0)$ to the line must be equal to the radius $c$.

From the equation of the line $x + y + 1 = 0$, we identify the coefficients: $A=1$, $B=1$, and $C=1$. Using the distance formula $d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$, and setting $d=c$: $c = \frac{|1(0) + 1(0) + 1|}{\sqrt{1^2 + 1^2}}$ Explanation: This is the core principle for determining if a line is tangent to a circle. The distance from the circle's center to the line must be exactly equal to the circle's radius. If this distance were less than the radius, the line would intersect the circle at two points; if it were greater, the line would not intersect the circle at all.

Step 5: Calculate the Value of c

Now, we perform the final calculation: $c = \frac{|0 + 0 + 1|}{\sqrt{1 + 1}}$ $c = \frac{|1|}{\sqrt{2}}$ $c = \frac{1}{\sqrt{2}}$

Thus, the value of $c$ is $\frac{1}{\sqrt{2}}$.

3. Important Tips and Common Pitfalls

• Memorize Standard Tangent Forms: Knowing the standard slope-form tangent equations for common conic sections (parabolas, ellipses, hyperbolas) is crucial for efficiency. Deriving them during an exam is time-consuming.
• Correctly Identify 'a': Be extremely careful to correctly identify the parameter 'a' for each parabola. While both were $a=1$ in this problem, they could easily be different in other questions, leading to different tangent equations.
• Careful with Signs: Pay close attention to signs, especially when solving equations like $m^3 = -1$. Remember that the cube root of a negative number is negative.
• Absolute Value in Distance Formula: Always include the absolute value in the numerator of the distance formula ($|Ax_0 + By_0 + C|$). Distance is a non-negative quantity. Forgetting this can lead to incorrect signs for $c$ (though $c$ here represents a radius and must be positive).
• Standard Form for Line: Ensure the line equation is in the $Ax+By+C=0$ form before applying the distance formula to correctly identify $A, B, C$.
• Real Roots for Slope: A physical tangent line in the Cartesian plane must have a real slope. If solving for $m$ yields only complex roots, it implies there is no real common tangent (or an error in calculation).

4. Summary and Key Takeaway

This problem serves as an excellent illustration of how to combine concepts from different conic sections. The methodical approach involved:

1. Expressing the general tangent equations for both parabolas in terms of a common slope $m$.
2. Equating the y-intercepts of these general tangent forms to determine the specific value of $m$ for the common tangent.
3. Substituting this common slope back into one of the general equations to find the unique equation of the common tangent line.
4. Finally, applying the fundamental tangency condition for a line to a circle (perpendicular distance from center equals radius) to solve for the unknown parameter $c$.

The key takeaway is the systematic application of standard formulas for tangents to conic sections and the distance formula for tangency conditions. These are fundamental tools in coordinate geometry and frequently appear in JEE problems.

The final answer is $\boxed{\frac{1}{\sqrt{2}}}$.