At what point do two lines extending from the end of a third equal each other? 
Depicted are three quarter circles with the same center point and different radii. The smaller circle has a radius of 2 (pink line), the mid of 4, and the largest of 5. From the end of the pink line, 2 perpendicular lines extend out. One (blue line) is bound by the quarter circle with radii 4. The other (red line) is bound by quarter circle of radii 5. As the pink line sweeps from 0 to 90 degrees, the blue line goes from its shortest length (2) to its longest length (3.46410161514) and the red line goes from its longest length (4.58257569496) to its shortest (3). My question is at what point in the pink lines sweep from 0 to 90 degrees does the length of the blue line and the length of the red line equal each other?
The question is aimed to answer a math problem from this video: https://youtu.be/iszucjbTh5I
The video offers a solution, but I haven't watched it and want to solve it differently.
My math knowledge is at a pre-calc level. I have good trig familiarity and I do a lot of recreational math.
My current intuition is to ascertain two functions that give a curve that represents the increase and decrease of the red and blue lines based on a degree value between 0 and 90. Then I'd set the two functions equal to each other and solve for sine of X. I could be way off in my approach but that is where I am at. Any help or suggestions would be greatly appreciated.
 A: 
We put the whole configuration on a Cartesian OXY Coordinate System, with the common center as the origin.
We know, any point on a circle with center as origin and radius $r$ can be described parametrically as $B(r\cos\theta, r\sin\theta)$. Here, $\theta\in\left[0,\dfrac{\pi}{2}\right]$ as only the quarter circle in the first quadrant is being considered. Here, $B(2\cos\theta, 2\sin\theta)$
Since $BC$ is horizontal, C will have the same ordinate as B, so that $C\equiv(x_C, 2\sin\theta)$. The equation of the orange circle is $$x^2+y^2=16.$$ so putting C in the equation, $$x_C=\sqrt{16-4\sin^2\theta}=2\sqrt{4-\sin^2\theta}$$ so the length of BC is $$|BC|= 2\sqrt{4-\sin^2\theta}-2\cos\theta$$
Similarly, $BD$ is vertical so D and B have same abscissa. Thus, $D\equiv (2\cos\theta, y_D)$. Equation of the red circle is $$x^2+y^2=25$$ so that $$y_D=\sqrt{25-4\cos^2\theta}= \sqrt{21+4\sin^2\theta} $$ so that $$|BD|=  \sqrt{21+4\sin^2\theta}-2\sin\theta.$$ Equating we get $$\left(2\sqrt{4-\sin^2\theta}-\sqrt{21+4\sin^2\theta}\right)^2=(2\cos\theta-2\sin\theta)^2$$ so that $$37-2 \sqrt{16-4\sin^2\theta} \sqrt{21+4\sin^2\theta}=4-4\sin2\theta$$$$\implies 33+4\sin2\theta= 2 \sqrt{16-4\sin^2\theta} \sqrt{21+4\sin^2\theta}$$ which on squaring again an simplifying, gives $$264\sin2\theta-72\cos2\theta=183$$ for which WolframAlpha gives two solutions.
A: Well, blue line is part of a chord in a circle with radius $4$, the red line is a part of a chord in a circle with the radius $5$. Let $x$ be the angle of the pink line. Then the length of the blue line is $b=\sqrt{16-4\sin^2 x}-2\cos x$ and the length of the red line is $r=\sqrt{25-4\cos^2 x}-2\sin x$So we get this equation to solve: $\sqrt{16-4\sin^2 x}-2\cos x=\sqrt{25-4\cos^2 x}-2\sin x$ 
I came up with $x\approx 1.33765$ or $76.64^\circ$.
A: 
Using Move slider in Geogebra for point B we can directly get adjusting approximate equal segments $(BC=3.03,BD=3.08)$ at $\angle BOX=78.25^{\circ}$
It can't get better pushing the mouse with the hand. Thanks to anyone helping with settings for a higher accuracy ( atleast 5 places).
A: 
Let the lengths of $BD$ and $BC$ be $x$. Then since $B = (2 \cos \theta, 2 \sin \theta)$ from the unit circle definition, and $BD, BC$ are parallel to the coordinate axes, $D = (2 \cos \theta, 2 \sin \theta + x)$ and $C = (2 \cos \theta + x, 2 \sin \theta)$.
Using the distance formula / Pythagoras given that we know $AD = 5, AC = 4$ results in:
$$(2 \cos \theta)^2 + (2 \sin \theta + x)^2 = 5^2$$
$$4 \cos^2 \theta + 4 \sin^2 \theta + 4x \sin \theta + x^2 = 25$$
$$4x \sin \theta + x^2 = 21 \tag{1}$$
and similarly:
$$(2 \cos \theta + x)^2 + (2 \sin \theta)^2 = 4^2$$
$$4 \cos^2 \theta + 4x \cos \theta + x^2 + 4 \sin^2 \theta = 16$$
$$4x \cos \theta + x^2 = 12 \tag{2}$$
Now in order to solve these equations, it would be easiest to isolate $\sin \theta, \cos \theta$ and then use the Pythagorean identity once more. Doing this gives after some rearrangement:
$$\left(\frac{21 - x^2}{4x} \right)^2 + \left(\frac{12 - x^2}{4x} \right)^2 = 1$$
$$(21 - x^2)^2 + (12 - x^2)^2 = (4x)^2$$
$$441 - 42x^2 + (x^2)^2 + 144 - 24x^2 + (x^2)^2 = 16x^2$$
and since everything is in terms of $x^2$, let's go ahead and substitute $x^2 = t$:
$$585 + (-66 - 16)x^2 + 2(x^2)^2 = 0$$
$$2t^2 - 82t + 585 = 0$$
We are getting close to the answer. Applying the quadratic formula:
$$t = \frac{41 \pm \sqrt{511}}{2} = x^2 \implies x \approx ±3.033, ±5.639$$
We should mention that $\theta \in [0, 2 \pi)$ before we continue further (look at the diagram to see what happens when $\theta$ is outside this domain). Now let us claim that each value of $x$ maps to exactly one value of $\theta$. Going back to equation $(1)$ for example and rearranging, $\sin \theta = \frac{21 - x^2}{4x}$. Now in this restricted domain, only $\pi - \theta$ will give us the same value of $\sin \theta$. However, $\cos(\pi - \theta)$ is $-\cos(\theta)$ through the unit circle, and since our $x$ is still the same, equation $(2)$ will not be true. Hence our conjecture is true and we have exactly $4$ possible candidates.
We can check that all $4$ candidates satisfy the system of equations we set up in equation $(1)$ and $(2)$. However, there are actually only $2$ solutions for $\theta$:

In short, this leaves $\theta \approx 1.337, 3.641 \ \text{rad}$ as the only two solutions where $BD = BC$, from which $(2 \cos \theta, 2 \sin \theta)$ will be the coordinates of point $B$.
I would encourage you to pause here and to play around with the setup on the GeoGebra visualisation here. Click on the spoilered text when you are ready:

 Rotating a point by $180$ degrees causes the signs of the lengths (in a Cartesian coordinate system) to change. However, when we square the lengths using Pythagoras, it doesn't matter if the signs are wrong.  (To make a further point, you can think of the horizontal and vertical components of $B$ and the length $x$ being in opposite directions in the wrong cases.)


 So there are only two solutions. It is very handy to consider the green line $y = x$ which makes an angle of $45$º with the $x$-axis. When $B$ is on the other side of the line as the two solutions, $BD > BC$ from the visualisation. You can try proving this by moving $B$ to one of two points where the green line intersects the circle with radius $2$, and then reflecting point $C$ across the line. Calling the reflected point $C'$, use a congruence theorem to prove that $\Delta C'BA \cong \Delta CBA$ and hence that $AC = AC'$, and so $AD$ must be even further than $AC' = AC$. (If you aren't convinced, consider a pair of right-angled triangles sharing the base on the green line, and for the last step, consider the x- and y-components of $AC'$ and $AD$). The two possible points are the two extreme cases: everywhere else you will find that $BD$ gets longer while $BC$ gets shorter by considering two equal halves of the region.



 Further clarification on "everywhere else you will find that $BD$ gets longer while $BC$ gets shorter by considering two equal halves of the region":


 

