Tangents of circles I'm trying to solve the following problem:
Find the tangent equations of $x^2 + y^2 = 1$ which pass though point $(1, 2)$.
As a line which goes though the point $(1, 2)$ is in the form $y = m(x - 1) + 2$ imagining $m$ as it's angular coefficient.
Making a substitution in the circle equation and finding it's determinant, it's roots in function of $m$ would give me the two angular coefficients of the tangent lines, but developing it, I can only get one equation:
$x^2 + (m(x - 1) + 2)^2 - 1 = 0$
Calculating the determinant:
$(-2m^2 + 4m)^2 - 4(1 + m^2)(m^2 - 4m + 3) = 0$
Which leads to:
$16m - 12 = 0$, $m = \frac{3}{4}$
Where's the other root ?

In order to find tangents to circles, the formula for the distance of a point to a line is better than substituting.
You want to find the lines through $(1,2)$ that have distance $1$ from the origin; the lines have equations
$$
ax+by=a+2b
$$
with $a$ and $b$ not both zero; the distance of this line from the origin is
$$
\frac{|a\cdot0+b\cdot0-a-2b|}{\sqrt{a^2+b^2}}
$$
so the equation you have to solve is
$$
(a+2b)^2=a^2+b^2
$$
that is
$$
4ab+3b^2=0
$$
that means $b=0$ or $4a+3b=0$. The coefficients are determined up to a multiplicative constant, so for the first you can choose $a=1$ and $b=0$; for the second you can choose $a=3$ and $b=-4$. Thus the two tangents are
$$
x=1\qquad\text{and}\qquad 3x-4y+5=0.
$$
