Tangents to parabola $y^2=-4(x-1)$ with slopes in a certain range determine chords, bisected by $x=1$, of a circle. Find the equation of the circle. Hi so I was doing this question

At first I didn't have the slightest clue on how to attempt it. So I decided to reverse engineer the problem. I looked up the answer and the answer was (a) so I plunged in the values in desmos and got this

Through this I got to saw that at m=0.5 and -0.5 the tangent at the parabola is also tangent at the circle and hence did the following calculation . 

In the last step I just compared the coefficients as shown just to get the answer. I got the answer but I am not sure about my last step. To be honest I am not even sure about the whole approach that I have taken. So can somebody please help get a better structured answer for this question.
 A: Your approach, though relevant, is not yet quite complete. You do need to show that any tangent of the parabola, which is also a chord of the circle, is bisected by the line $x = 1$ as required by the question. Here is one way to do so.

Assume a point on the parabola $y^2=-4(x-1)$, say,
$$\left( b, 1-\frac{b^2}{4} \right) $$
Because for the parabola, $x'= -y/2$, the slope of the tangent line at this point is,
$$ k_1 = -\frac{b}{2} $$
and the tangent line is give by
$$x-1+\frac{b^2}{4}= -\frac{b}{2}(y-b)$$
It intersects with the line $x = 1$ at the point,
$$\left(\frac{b}{2}, 1\right) $$
Then, the slope of the line passing through this point and the center of the circle is simply,
$$k_2 = \frac{2}{b}$$
Observing that the two slopes satisfy,
$$k_1 = -\frac{1}{k_2} $$
which means that they are perpendicular to each others. Thus, the line of $k_2$ is also a radius line, hence bisecting the chord.
A: This problem is somewhat difficult to grasp, especially without graphical tools. Here's a bottom up approach. The main ingredient necessary to understand this problem is that, given a circle of a given radius centered at the origin, a subset of its chords can be mapped onto tangents of a certain subset of the parabola. The circle has to be centered at the origin for the given parabola, or the trick doesn't work. Conversely, restricting the slopes of the tangents to a certain interval$($here $\mathbb{R}-(-1/2,1/2))$ results in a certain radius the circle that the chords bisected by $x=1$ form must have.
Pick an arbitrary point of the parabola $P(x_0,y_0)$. The tangent to the parabola through this point is described by the equation:
$$y-y_0=-\frac{2}{y_0}(x-x_0)$$
The line $x=1$ hits the tangent at the point $M(1,\frac{y_0}{2})$. This is the middle point of the chord. We do not know what the chord's length is, but we understand that it depends on the point $P$ chosen, so let us name the length of the chord $D(y_0)$. Since the point $M$ is right in the middle of the chord, then if we move along the tangent by $D/2$ to the right or to the left we should be on the circle. If $\mathbf{t},\mathbf{r}_L, \mathbf{r}_R, \mathbf{r}_0$ are respectively the normalized vector parallel to the tangent lines pointing to the right, the coordinates of the leftmost and rightmost points of the chord, and the coordinates of $M$ we obtain
$$\mathbf{r}_L=\mathbf{r}_0-\frac{D}{2}\mathbf{t}\\\mathbf{r}_R=\mathbf{r}_0+\frac{D}{2}\mathbf{t}\\\mathbf{t}=\frac{|y_0|}{\sqrt{y_0^2+4}}(1,-\frac{2}{y_0})$$
We stipulate that the leftmost and rightmost points of the chord trace a circle of radius $R$ centered at $\mathbf{c}$ and therefore for all values of $y_0$ we must have:
$$||\mathbf{r}_L-\mathbf{c}||=R~~,~~||\mathbf{r}_R-\mathbf{c}||=R$$
We note that:
$$\mathbf{r}_L+\mathbf{r}_R=2\mathbf{r}_0 ~~,~~ \mathbf{r}_R-\mathbf{r}_L=D\mathbf{t}$$
Square both equations and subtract to obtain after some algebra the equations:
$$\mathbf{t}\cdot\mathbf{r}_0=\mathbf{t}\cdot\mathbf{c}\iff c_x y_0-2c_y=0 ~\forall~y_0\Rightarrow \mathbf{c}=0$$
which fixes the circle to be at the origin. Now by some simple trigonometry, the distance of $M$ from the origin (which happens to be $||\mathbf{r}_0||$) can be related to the  length of the chord and the radius of the circle as follows:
$$\mathbf{r}_0^2=R^2-\frac{D^2}{4}$$
which yields 
$$D(y_0)=\sqrt{4(R^2-1)-y_0^2}$$
Suppose that we only allow part of the parabola as discussed above, $|y_0|\leq y_c$. Then all the tangents to the points allowed belong to the circle as long as $D(y_c)=0$. This last condition determines the radius to be:
$$R=\sqrt{1+\frac{y_c^2}{4}}=\sqrt{1+\frac{1}{m_{min}^2}}$$
where $m_{min}$ is the minimum slope allowed. In our problem, $m_{min}=1/2$ and the radius is $R=\sqrt{5}$ as required.
A: You should be able to proceed step by step from the given conditions instead of having to work backwards from the solution.  
To start off with, find the an equation for the tangent with slope $m$ to the parabola. There are various ways to do this. I’ll use homogeneous matrices and the dual conic to the parabola. The matrix of this parabola is $$C = \begin{bmatrix}0&0&2\\0&1&0\\2&0&-4\end{bmatrix}$$ and for the line $\mathbf l = (m,-1,b)^T$ to be tangent to this we must have $\mathbf l^TC^{-1}\mathbf l = m^2+mb+1 = 0$, from which $b=-(1+m^2)/m$. This intersects the line $x=1$ at the point $M$ with (inhomogeneous) Cartesian coordinates $(1,-1/m)^T$.  
Now, the center of a circle lies on the perpendicular bisector of every chord. We’re given that $M$ is always the midpoint of a chord of the circle. The perpendicular to $\mathbf l$ through $M$ works out to be $(1,m,0)^T$, i.e., $x+my=0$. This line passes through the origin for any value of $m$, therefore the circle is centered at the origin. This happens to be the parabola’s focus.  
When the line $y=mx-(1+m^2)/m$ has two distinct real intersections with the circle $x^2+y^2=r^2$, it turns out that they are symmetric with respect to $M$ (verify this!). In other words, if any tangent to the parabola is secant to a circle centered at the origin, then the midpoint of the chord lies on the line $x=1$ regardless of the circle’s radius. The wording of the problem is somewhat unfortunate. It would appear that “All the tangents to the parabola ... whose slopes lie in the interval ...” is meant to be read as saying that only the tangents with slopes in $\mathbb R\setminus[-1/2,1/2]$ are secant to the circle. Otherwise, the best we can do is give a minimum radius.  
With that reading, the tangents with slope equal to $\pm1/2$ are tangent to the circle. Using the standard point-line distance formula, we can find that the square of the distance of the tangent line $\mathbf l$ from the origin is equal to $(1+m^2)/m^2$, and setting $m=1/2$ we get $\sqrt5$ for the circle’s radius. Equivalently, we note that $M$ is the point of tangency to the circle for these two lines and compute its distance from the origin.
