A tangent line $$L$$ is drawn at the point $$(2, -4)$$ on the parabola $$y^2 = 8x$$. If the line $$L$$ is also tangent to the circle $$x^2 + y^2 = a$$, then $$a$$ is equal to _________.

We begin with the parabola whose equation is $$y^{2}=8x$$. In the standard form $$y^{2}=4ax$$ we identify $$4a=8$$, so we get $$a=2$$ for the parameter of the parabola.

For a parabola of the type $$y^{2}=4ax$$, the equation of the tangent at a point $$(x_{1},y_{1})$$ on the curve is given by the standard tangent formula

$$yy_{1}=2a\left(x+x_{1}\right).$$

Here the given point is $$(2,-4)$$, so we have $$x_{1}=2$$ and $$y_{1}=-4$$. Substituting these values along with $$a=2$$ into the tangent formula we obtain

$$y(-4)=2\cdot2\,(x+2).$$

Simplifying the right-hand side first, $$2\cdot2=4$$, so

$$-4y=4(x+2).$$

Next we divide every term by $$-4$$ to get the equation in a neater form:

$$y=-x-2.$$

It is often convenient to write a straight-line equation in the form $$Ax+By+C=0$$. Adding $$x$$ and $$2$$ to both sides gives

$$x+y+2=0.$$

Thus the tangent line $$L$$ has the Cartesian equation $$x+y+2=0.$$ The same line is also tangent to the circle whose equation is $$x^{2}+y^{2}=a.$$ The centre of this circle is clearly the origin $$(0,0)$$ and its radius is $$\sqrt{a}.$$

For a line $$Ax+By+C=0$$, the perpendicular distance of a point $$(x_{0},y_{0})$$ from the line is given by the distance formula

$$\text{Distance}=\dfrac{|Ax_{0}+By_{0}+C|}{\sqrt{A^{2}+B^{2}}}.$$

To enforce tangency, the distance from the centre of the circle to the line must be equal to the radius. Taking $$(x_{0},y_{0})=(0,0)$$ and the coefficients $$A=1,\;B=1,\;C=2,$$ we compute

$$\text{Distance}=\dfrac{|1\cdot0+1\cdot0+2|}{\sqrt{1^{2}+1^{2}}} =\dfrac{|2|}{\sqrt{2}} =\dfrac{2}{\sqrt{2}} =\sqrt{2}.$$

This distance must equal the radius $$\sqrt{a},$$ so we set

$$\sqrt{a}=\sqrt{2}.$$

Squaring both sides gives

$$a=2.$$

So, the answer is $$2$$.