Tangents from a certain point, to a circle? 
We have a circle with radius 2, centred on the origin. Find the equation of the lines passing through the point $(0,4)$ which are tangent to the circle.

So we have the circle $$x^2 + y^2 = 4$$
We need 2 lines $$y=ax+4$$
So if you fill that in the equation of the circle, we end up with $$a^2x^2 + 8ax + x^2 + 12 = 0 $$.
So we'd have to solve $$ a = \dfrac{-8x \pm \sqrt{64x^2-4x^2(x^2+12)}}{2x^2}$$
I must have done something wrong, since this causes the discriminant to be negative, and thus gives no answer. Is there a smarter way I overlooked?
 A: Suppose $A(x_1,y_1)$ is the point of the circle where the tangent is drawn.
Then the equation of the tangent is $xx_1+yy_1=4$. Since $(0,4)$ belongs to the tangent it satisfies her equation so $0x_1+4y_1=4\rightarrow y_1=1$.
Then subtitute this value to the circle equation 
we get $x_1^2+y_1^2=4\rightarrow x_1^2+1^2=4\rightarrow x_1=\sqrt3$ or $x_1=-\sqrt3$.
Thus we have two tangents with equations:
$\sqrt3x+y=4$ and $-\sqrt3x+y=4$.

A: The values of $x$ represents the abscissa of the intersection 
The equation will be $$(a^2+1)x^2+8ax+12=0$$ 
For tangency, the roots of the above quadratic Equation must be same, hence the discriminant must be $0$ 

Alternatively,
The equation of any line passing through $(0,4)$ will be $\frac{y-4}{x-0}=m\iff mx-y+4=0 $ where $m$ is the gradient
Now, the perpendicular distance of a tangent from the center $(0,0)$ of the circle will be $=$ radius$(=2)$ 
A: You can get the tangent points by doing the intersection between the circle and the circle of radius $2$ centered in $(2,0)$ (the lines from one of this points to $(0,0)$ and from the same point to $(4,0)$ will be perpendicular)
so you have 
$$x^2 + y^2 = 4$$
and
$$(x-2)^2 + y^2 = 4$$
The x coordinate will be the same for the two points, so
$$x^2 - 4 = (x-2)^2 -4$$
from here you get $x=1$ and $y^2=3$
So you have the two points of tangency, and from here you can easily find the equations of the tangents.
