Solution

Let the common tangent be

$y=mx+c$

For the parabola

$y=-x^2$

we have

$-x^2=mx+c$

$x^2+mx+c=0$

For tangency, discriminant must be zero.

$m^2-4c=0$

So,

$c=\frac{m^2}{4}$ .....$(1)$

Now, for the parabola

$y=(x-2)^2$

we have

$(x-2)^2=mx+c$

$x^2-4x+4=mx+c$

$x^2-(m+4)x+(4-c)=0$

For tangency,

$(m+4)^2-4(4-c)=0$

Substitute $c=\frac{m^2}{4}$:

$(m+4)^2-16+4\cdot \frac{m^2}{4}=0$

$(m+4)^2-16+m^2=0$

$m^2+8m+16-16+m^2=0$

$2m^2+8m=0$

$2m(m+4)=0$

So,

$m=0$ or $m=-4$

From the given options, $m=-4$ is possible.

Using $(1)$,

$c=\frac{(-4)^2}{4}$

$c=4$

So, the common tangent is

$y=-4x+4$