回答所有问题。

1. Substitute the equation of the line into the equation of the curve

Substitute $y = kx - 2$ into $y^2 = 4x - x^2$:
$(kx - 2)^2 = 4x - x^2$

2. Expand and rearrange into a standard quadratic equation in $x$

$k^2x^2 - 4kx + 4 = 4x - x^2$
$k^2x^2 + x^2 - 4kx - 4x + 4 = 0$
$(k^2 + 1)x^2 - (4k + 4)x + 4 = 0$

3. Apply the discriminant condition for the line to meet the curve

For the line to meet the curve, the quadratic equation must have real roots, so $b^2 - 4ac \geqslant 0$.
$[-(4k + 4)]^2 - 4(k^2 + 1)(4) \geqslant 0$

4. Solve the inequality for $k$

$16(k + 1)^2 - 16(k^2 + 1) \geqslant 0$
Divide by 16:
$(k^2 + 2k + 1) - (k^2 + 1) \geqslant 0$
$2k \geqslant 0$
$k \geqslant 0$

1. Apply the discriminant condition for a quadratic expression to be always positive.

For $2x^2 + bx + c > 0$ for all real $x$, since the coefficient of $x^2$ is $2 > 0$, the discriminant must be less than zero.
$b^2 - 4(2)(c) < 0$

2. Simplify the inequality to find the relationship between $b$ and $c$.

$b^2 - 8c < 0$
$b^2 < 8c$

3. Explain why $c$ is never negative.

Since $b$ is a real number, its square is non-negative, so $b^2 \ge 0$.
Given $8c > b^2$, we have $8c > 0$, which implies $c > 0$.
Thus, $c$ is strictly positive and can never be negative.
(Alternatively, substituting $x=0$ into $2x^2+bx+c$ gives $c$. Since the expression is always positive, $c > 0$.)