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 ?

It should be immediately apparent that because the radius of the circle is $1$ and the point through which the tangent line must pass is $(1,2)$, that the second tangent line is $x = 1$, with infinite slope.  That is why you only got one solution using your method; you assumed a particular form for the tangent line, and one of the tangent lines does not satisfy this form.
