<s>[INST] 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?
 [/INST] A possible approach:
1) Calculate distance from (0,0) to line $y = mx + c$. 
The perpendicular line passing through the origin is 
$y = - (1/m)x$.
Point of intersection:  $- (1/m)x = mx + c$.
$- x = m^2x + mc$
$x ( m^2 + 1) + mc = 0$;
$x = (-mc)/(m^2 +1)$;
Corresponding $y$:  
$y = - (1/m) (-mc) /(m^2 + 1)$;
$y = c/(m^2 +1)$.
Distance (squared) to the origin:
$x^2 + y^2 = [(mc)^2 + c^2]/(m^2 +1)^2$ = 
=  $c^2(m^2 +1)/(m^2 + 1)^2 $= 
$c^2/(m^2 + 1)$.
2) For this point to be a point of tangency:
Distance (squared) = radius (squared) =
$a^2$.
Putting together:
$c^2/(m^2 +1) = a^2$.
Answer 2).
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