Hey there!

When jnlfs2010 said this:

You've got a quadratic. Now get into ax^{2}+bx+c=0 form. Since the line is a tangent, the discriminant of that equation must equal to 0 as there is only one solution. With the value of m, you surely should get the formula of the tangent

They essentially implied expanding and simplifying the quadratic you ended up with, which was \(x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0\), which also happened to be book's solution. They also told you to use the discriminant, and let it equal to zero.

Essentially, when a line cuts the parabola, you have a maximum of two solutions, and a minimum of zero. Equating the line and the parabola will give you the x-values of the points of intersection as roots of a quadratic. If the quadratic is a perfect square, then \(\Delta = 0\), and there is one real root ie. the line is a tangent to the parabola, and does not in fact cut the parabola in two different places like when the discriminant is greater than zero ie. the 'two roots' are equal - there is thus only one real root!

Since we want to find the tangent to the circle at (2, 1), there is only one solution (or if you want to think about it like that, there are two equal solutions!) since there is only one point of contact. Hence, the book (and we!) choose to expand the brackets in this quadratic in x \(x^2 +(mx-2m+1)^2 - 4(mx-2m+1) -1=0\) , then find the discriminant using \(\Delta = b^2 - 4ac\), and let it equal zero, since we only want the one solution. They've got in the book that b

^{2} = 4ac, which is the same as b

^{2} - 4ac equals zero. Since there is only one solution, we expect a perfect square with a quadratic in m, thus finding the gradient of the tangent, then using that to find the equation of the line using point gradient form.

A quicker alternative method if you've learned it is suggested by ^^^111^^^, implicit differentiation. Implicit differentiation is just an extension of the chain rule, ie.

In our case, we have that

Isolating \(\frac{dy}{dx}\) and substituting the coordinates of any point on the given circle will give the gradient of the tangent at that point, just like differentiating any other curve. From here, the steps are the same as the method given in the book.

Hope this clears everything up!