Equation of a tangent that passes through a point not on the curve Just made this little question for me and a friend and we couldn't find the answer given what we know, which is only fairly basic calculus, similar to what you would learn in the first year of 'A level' maths.

Find the equation of the line which is a tangent to the curve $x^2-2x-1$ and passes through the point $(5,5)$. 

I can find the gradient of the curve at $x$ of $2x-2$ fairly easily, and I know that the line $y=mx+c$ will have a gradient $m$ which is equal to $2x-2$ at the point where they intersect, but I don't know how to form an equation given this.
Any simple explanation of how to answer this would be greatly appreciated.
Here is a visualisation of the problem:
Image captured on an graphing calculator
The red curve is the quadratic I described and the blue is the tangent I'm looking for.
Thanks in advance.
 A: Hint:
Using your approach, note that ,if $Q=(x_Q,y_Q)$ is the point of tangency than the slope of the line is:
$$
m=\frac{y_Q-5}{x_Q-5}
$$
Since $Q$ is point of the parabola, we have $y_Q=x_Q^2-2x_Q-1$, and we want that this slope is the same as the derivative of the function at $x_Q$, so we have the equation:
$$
\frac{f(x_Q)-5}{x_Q-5}=f'(x_Q)
$$
that is:
$$
\frac{x_Q^2-2x_Q-1-5}{x_Q-5}=2x_Q-2
$$
solve this equation and find $x_Q$, so you can find the point $Q$ and the slope of the tangent.
But you  can use also another approach that does not require derivatives.
The lines passing for the given point are:
$$
y-5=m(x-5)
$$
and we want the line that is tangent to the parabola, this means that this line has only one common point with the parabola, so, write the system:
$$
\begin {cases}
y-5=m(x-5)\\
y=x^2-2x-1
\end{cases}
$$
and find for what values of $m$ it has only one solution, i.e. for what $m$ the discriminant of the system is equal to 0. This is the slope of the tangent.       
A: HINT.
$$
{y-5\over x-5}=m=2x-2.
$$
A: You should try and understand what the equation of a line means. Consider the following:
$$y - y_1 = m(x - x_1)$$
What this equation is telling you is that for every movement of $(x - x_1)$ on the x-axis we go up or down by $(y - y_1)$ on the y-axis. The value $m$ (i.e. the slope) is like a tuning variable that controls how quickly you go up or down and you can obtain this value using the derivative. 
If you understand this then solving the problem is pretty simple. Just note that you can obtain the point $(x_1, y_1)$ using the original equation
