Linear equation with circle What is the equation of a line that is tangent to the circle with center at the origin and a radius of 1 and passes through the point through the point (0,2)? 
I tried it for quite a while but still can't find a straightforward method of doing this problem.
The answer is y=-rad3*X+2
 A: Let $ax+by+c=0$ be the equation of the line.
the distance from the center to this tangent line is the radius.
$$\frac {|c|}{\sqrt {a^2+b^2}}=1$$
the point $(0,2) $ belongs to this line, so
$$2b+c=0\;\;,\;\;c=-2b$$
from this
$$a^2+b^2=4b^2\;\;,\;\;a=\pm b \sqrt {3}$$
the equation is then

$$\pm x\sqrt {3}+y=2$$

it is normal to find two lines since the point $(0,2) $ is in a symetry axis.
A: Let $A(0,2)$,  $C$ be a tangency point and $O$ be an origin.
Hence, $\measuredangle ACO=90^{\circ}$ and since $OC=\frac{1}{2}AO$, we get $\measuredangle OAC=30^{\circ}$.
Thus, $m_{AC}=\pm\sqrt3$ and we get the answer:
$$y-2=\pm\sqrt3x.$$
Done!
A: Here is a figure from trigonometry that is worth noting.

if $\csc t = 2$ then $t = \frac {\pi}{6}$ and $\sec t = \frac {2}{\sqrt 3}$
Intercept-intercept equation of a line.
$\frac {x}{a} + \frac {y}{b} = 1$
and with these intercepts.
$\frac {x \sqrt 3}{2} + \frac {y}{2} = 1$
And of course it can be reflected across the y axis.
$\frac {x \sqrt 3}{2} - \frac {y}{2} = 1$
