The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if: The line $y=mx+c$ is a tangent to $x^2+y^2=a^2$ if:
$1$. $c=a\sqrt {1+m^2}$
$2$. $c=\pm a\sqrt {1+m^2}$
$3$. $c^2=\pm a\sqrt {1+m^2}$
$4$. $\textrm {None}$
My Attempt:
The tangent to circle $x^2+y^2=a^2$ at point $(x_1,y_1)$ is given by :
$$xx_1+yy_1=a^2$$
Now, what should I do next?

You have $x^2 + (mx+c)^2 = a^2 \iff x^2 + m^2x^2 + 2mxc + c^2 = a^2$ which simplifies to $$(1+m^2)x^2 + 2mcx + (c^2-a^2) = 0$$
But as $y$ must be tangent, there must be a single solution to the above quadratic. Which means its discriminant must be $0$, i.e: $$4m^2c^2-4(1+m^2)(c^2-a^2) = 0$$
