Stuck with a possibly impossible trigonometry question

Question

I need to find the length of the arc between Y1 and Z1 in the image below. If you can even get me to the value of Y, then that will work.

I appreciate the drawing may be crude, but imagine that Y and Y1 are directly above each other, as are C, Z, Z1, and the line with A, X, Y is perfectly horizontal (i.e. perpendicular to the Y and Z lines mentioned before)

I have no idea if this is even possible, and I've been stuck for over an hour so am throwing it out there to anyone who has better than High School level math than I do!

UPDATE: I have been told I should look into using quadratic equations to find the two intersections of a line between a circle, and I think that should be enough to get me there.

In case there's another way, I'll add some more info.

This is actually based on coords of pixels on a screen. I know the coords of A, C, X, Z and Z1. I also know the lengths of R, AZ and AZ1. The only value that can change is X, although the more I look at it, the more I think it's irrelevant to my situation i.e. finding the length of the arc Y1Z1.

Answer 1

User141288 provided the best possible solution. However, it is also possible to solve this problem in a trivial way by using analytical geometry.

You say that you know the coords of $A(x_A,y_A)$, $C(x_C, y_C)$ and $Z1(x_{Z1}, y_{Z1})$.

So you can easily calculate the radius of the circle:

$$R^2=(x_{Z1}-x_C)^2+(y_{Z1}-z_C)^2$$

Equation of straight line passing through points $A,Z_1$ is:

$$y=y_A+\frac{y_{Z1}-y_{A}}{x_{Z1}-x_{A}}(x-x_A)\tag{1}$$

On the other side, the equation of circle is:

$$(x-x_C)^2+(y-y_C)^2=R^2$$

$$(x-x_C)^2+(y-y_C)^2=(x_{Z1}-x_C)^2+(y_{Z1}-y_C)^2\tag{2}$$

The point is: Equations (1) and (2) have two unknowns ($x$,$y$). Replace (1) into (2) and you will get a single quadratic equation that can be solved for $x$. In a general case that equation (like any other quadratic equation) has zero, one or two different real solutions. Assuming that your solutions are both real and different $(x_1, x_2)$, you can now calculate $(y_1,y_2)$ from (1).

The distance between intersection points (and note that one of them has to be $Z_1$ if you calculated everything correctly) is simply:

$$Y_1Z_1=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Answer 2

I suppose "directly above" means the points have the same $x$ coordinate on the screen and "perfectly horizontal" means the points have the same $y$ coordinate on the screen.

The triangle $\triangle CY_1Z_1$ is isosceles. If $M$ is the midpoint of the side $Y_1Z_1$, then the line $CM$ cuts the isosceles triangle into two right triangles, $CMY_1$ and $CMZ_1$. In these triangles, $\angle MCY_1 = \angle MCZ_1.$

The question implies that the line $CZZ_1$ is perpendicular to the line $AYZ.$ Because of that and because the line $CM$ is perpendicular to the line $AY_1Z_1,$ the angles $\angle ZAZ_1$ and $\angle MCZ_1$ are congruent.

The length of the arc is simply $R$ times the radian measure of the angle $\angle Z_1CY_1.$ But $$\angle Z_1CY_1 = 2 \angle MCZ_1 = 2\angle ZAZ_1. $$

So you just need to find $\angle ZAZ_1,$ double it, make sure it's in radians, and multiply by $R.$

There are various choices for computing $\angle ZAZ_1.$ You could take $\angle ZAZ_1 = \arccos(AZ/AZ_1),$ though in the figure as shown, $\angle ZAZ_1 = \arcsin(ZZ_1/AZ_1)$ would tend to give a more accurate result. You could also take $\angle ZAZ_1 = \arctan(ZZ_1/AZ).$

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In case you need the straight-line distance between $Y_1$ and $Z_1,$ that's even simpler: using similarity of the three right triangles $\triangle CMZ_1$, $\triangle CMY_1$, and $\triangle AZZ_1$, $$ Y_1Z_1 = 2R \frac{ZZ_1}{AZ}.$$