Question

Sol. [Key Concept: A common tangent $y = mx + c$ has $c = \frac{10}{m}$ (parabola) and $c^{2} = 50(1 + m^{2})$ (circle); solve the pair for $m,c$.]

A tangent to $y^{2} = 40x\ (a = 10)$ is $y = mx + \frac{10}{m}$;

tangency to $x^{2} + y^{2} = 50$ needs $\frac{|10/m|}{\sqrt{1 + m^{2}}} = \sqrt{50}$.

Then $100 = 50\ m^{2}\left( 1 + m^{2} \right)$

$\Rightarrow \ m^{4} + m^{2} - 2 = 0 \Rightarrow \ m^{2} = 1$, so $m = \pm 1$:

the common tangents are $x - y + 10 = 0$ and $x + y + 10 = 0$ → (b), (c).