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 ?
 A: 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.
A: 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
$$
x=1\qquad\text{and}\qquad 3x-4y+5=0.
$$
A: Note on a rather different approach or two (draw a diagram).
It is obvious from a diagram that $x=1$ is a tangent (and the proof is trivial). The centre of the circle is $(0,0)$ - exploit symmetry and reflect the tangent in the line $y=2x$ thorough the centre and the point you have been given to get the other one.
Alternatively if $x^2+y^2=1$ then differentiating with respect to $x$ we get $2x+2yy'=0$, or $y'=-\frac x y$
So the tangent to the circle through the point $(x_1,y_1)$ is the line $$y-y_1=-\frac {x_1}{y_1}(x-x_1)$$ or $$yy_1+xx_1=x_1^2+y_1^2=1$$
Note this is a fairly standard kind of form for a tangent to a circle, with the origin adjusted as necessary. Worth knowing if you are going to be tackling more of these.
This passes through $(1,2)$ provided $2y_1+x_1=1$ or $x_1=1-2y_1$
Substitute in the equation for the circle to obtain $$y_1^2+1-4y_1+4y_1^2=1$$0r $$y_1(5y_1-4)=0$$
I mention this because it seems to involve simpler calculations and less scope for error. Note that the symmetric form of the equation for the tangent has no division by zero, whereas working with one variable (e.g. the form of the derivative obtained by division) can introduce a division by zero and lose a solution. It is possible to do all the calculations without a potential division by zero and this is worth noting. Above all draw a diagram and exploit obvious symmetry if you can..
