If the common tangent to the parabolas, $$y^2 = 4x$$ and $$x^2 = 4y$$ also touches the circle, $$x^2 + y^2 = c^2$$, then $$c$$ is equal to:

1. Finding the Common Tangent

The equation of a tangent of slope $$m$$

to the parabola  $$y^2 = 4ax$$ is given by:

$$y = mx + \frac{a}{m}$$

For the given parabola $$y^2 = 4x$$, we have $$a = 1$$

. Therefore, the tangent is:

$$y = mx + \frac{1}{m}$$

The equation of a tangent of slope $$m$$

to the parabola $$x^2 = 4ay$$ is given by:

$$y = mx - am^2$$

For the given parabola $$x^2 = 4y$$ , we have $$a = 1$$.

 Therefore, the tangent is: $$y = mx - m^2$$

For a line to be a common tangent to both parabolas, the equations must be identical. This means their y-intercepts must be equal:

$$\frac{1}{m} = -m^2$$

$$m^3 = -1$$

$$m = -1$$

Substitute

$$m = -1$$

back into the tangent equation to get the equation of the common tangent:

$$y = (-1)x + \frac{1}{-1}$$

$$y = -x - 1$$

$$x + y + 1 = 0$$

2. Condition for Tangency to the Circle

We are given that the common tangent $$x + y + 1 = 0$$

also touches the circle: $$x^2 + y^2 = c^2$$

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.

The center of the circle

$$x^2 + y^2 = c^2$$

is at $$(0, 0)$$ and its radius is $$c$$.

Using the perpendicular distance formula from a point

$$(x_1, y_1)$$ to a line $$Ax + By + C = 0$$:

$$\text{Distance} = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Substitute the center

$$(0, 0)$$

and the line

$$x + y + 1 = 0$$

into the formula, and set it equal to the radius

$$c$$

:

$$c = \frac{|1(0) + 1(0) + 1|}{\sqrt{1^2 + 1^2}}$$

$$c = \frac{|0 + 0 + 1|}{\sqrt{1 + 1}}$$

$$c = \frac{1}{\sqrt{2}}$$

Final Answer

The correct value of

$$c$$

is:

$$c = \frac{1}{\sqrt{2}}$$

This corresponds to Option B.