Complex number problem

[Aion]

Problem:

Among the complex numbers $z$ which satisfy the condition $\lvert z-25i\rvert \leq15$ find the number having the least positive argument.

Solution attempt.

The condition $\lvert z - 25i \rvert \leq 15$ describes all complex numbers $z$ that lie within or on the boundary of a circle centered at the complex number $25i$, with a radius of $15$. Geometrically, this corresponds to the set of points in the complex plane whose distance from $25i$ is less than or equal to $15$. Our goal is to determine the complex number $z$ within this region that has the smallest positive argument.

The argument of a complex number is minimized when the angle $\alpha$ it makes with the positive real axis is as small as possible, while still being positive. In this case, this occurs at the point in or on the circle that lies furthest to the right—that is, the point closest to the real axis in the first quadrant. This specific point lies on the boundary of the circle and corresponds to the smallest positive angle $\alpha$ such that a ray emanating from the origin at angle $\alpha$ is tangent to the circle. In other words, it is the point of tangency between the circle with radius $15$ and centre $(0,25)$ and a ray from the origin in the first quadrant.


The figure above is a geometrical interpretation of the problem. We are asked to find the complex number C, which is the point of tangency of the straight line $$y=mx=tan(\alpha) x$$ with the circle $$x^2+(y-25)^2=15^2$$ as $\alpha$ is the smallest positive angle.

For the line $y=mx$ to be tangent to the circle, the distance from the centre $(0,25)$ to the line $y=mx$ must equal the radius $15$. From the distance formula in coordinate geometry, we know that the length of the perpendicular from the point $(x',y')$ to the straight line $Ax+By+C =0$ is given by
$$\lvert\frac{Ax'+By'+C}{\sqrt{A^2+B^2}} \rvert$$
In this case from $mx-y=0$ we have $A=m,B=-1$ and $C=0$. Hence, the distance from the straight line $mx-y=0$ and the centre $(0,25)$ is $15$. This gives us

$$R=\frac{\lvert m(0)-1(25)+0\rvert}{\sqrt{m^2+1}}=\frac{25}{\sqrt{m^2+1}}=15$$
Solving for $m$ we obtain $m=\pm\frac{4}{3}$. Since we are only interested in the solution in the first quadrant, we conclude that the equation for the tangent line is $y=\frac{4x}{3}$ and we note that $\alpha=arctan(\frac{4}{3})$

By substituting $y=\frac{4x}{3}$ into the equation for the circle $x^2+(y-25)^2=15^2$ we have the following

$$x^2+\left(\frac{4x}{3}-25 \right)^2=225$$$$25x^2-600x+3600=0$$$$x^2-24x+144=0$$$$(x-12)^2=0$$Therefore $$x=12, y=\frac{4}{3}\cdot12=16$$
So the point of tangency is $$C=12+16i$$
Could you review my solution? Is there a simpler approach to arrive at the same result? Also, is my method correct? Thanks!

 

[Dr.Peterson]

I can solve it geometrically in my head; is that accepted?

Since angle alpha = angle OAC, and cos(alpha) = cos(OAC) = 15/25 = 3/5. So x = cos(alpha)*OC = 3/5*20 = 12; similarly, y = sin(alpha)*OC4/5*20 = 16. QED.

 

[blamocur]

Looks good to me. A simpler geometric solution would be to use similar triangles.

 

[Aion]

I can solve it geometrically in my head; is that accepted?

Since angle alpha = angle OAC, and cos(alpha) = cos(OAC) = 15/25 = 3/5. So x = cos(alpha)*OC = 3/5*20 = 12; similarly, y = sin(alpha)*OC4/5*20 = 16. QED.

This is a remarkably straightforward solution. Upon seeing it, I realize that my own approach was unnecessarily complex.