Animated Solution for Mathematics - Complex Numbers: The number of points of intersection of $|z-(4+3i)|=2$ and $|z|+|z-4|=6,z\in \mathbb{C}$ is :

A Dance of Curves

Welcome, future engineer! Today, we are not just solving an equation; we are embarking on a journey into the complex plane.

When you see $|z-(4+3i)|=2$ and $|z|+|z-4|=6$, don't panic. Instead, see them as two elegant dancers moving across the Argand plane. Our goal is to find where they meet.

The Circle

Let's look at the first equation: $|z-(4+3i)|=2$. This is the classic form $|z-z_0|=r$.

In the complex plane, this is the definition of a circle. The center $z_0$ is $4+3i$, which corresponds to the Cartesian point $(4,3)$. The radius $r$ is $2$.

Imagine a circle centered at $(4,3)$ with a radius of $2$. It is a simple, perfect shape, but where does it sit relative to our second curve?

The Ellipse

Now, consider the second equation: $|z|+|z-4|=6$. This is the definition of an ellipse.

The sum of the distances from any point $z$ to two fixed points (the foci) is constant. Here, the foci are $z_1=0$ (the origin) and $z_2=4$ (the point $(4,0)$). The constant sum is $2a=6$, so $a=3$.

The center of the ellipse is the midpoint of the foci:

To draw this, we need the semi-minor axis $b$. The distance between the foci is $2ae=4$, so $ae=2$.

Using the fundamental relation $b^2=a^2-(ae{)}^2$, we find:

Thus, $b=\sqrt{5}\approx 2.236$. Our ellipse is centered at $(2,0)$ with a major axis of $3$ and a minor axis of $\sqrt{5}$.

The Collision

This is where the magic happens. We have a circle and an ellipse. Do they intersect? Let's test the vertical line $x=4$.

For the circle, the equation is $(x-4{)}^2+(y-3{)}^2=4$. At $x=4$, this becomes $(4-4{)}^2+(y-3{)}^2=4$, which simplifies to $(y-3{)}^2=4$.

This gives $y-3=\pm 2$, so $y=5$ or $y=1$. The lowest point of the circle on the line $x=4$ is $y=1$.

Now, let's look at the ellipse:

At $x=4$, we have:

Solving for $y^2$, we get:

This means $y=\pm \frac{5}{3}\approx \pm 1.67$.

At $x=4$, the ellipse reaches a height of $1.67$. Since the circle's lowest point is at $y=1$, which is below the ellipse's top edge of $1.67$, the circle must dip inside the ellipse!

Because the center of the circle $(4,3)$ lies outside the ellipse (since $3>\sqrt{5}$), and the circle dips inside the ellipse at $x=4$, it must cross the boundary of the ellipse twice to enter and exit.

Therefore, there are exactly 2 points of intersection. You have successfully navigated the complex plane using nothing but geometric intuition and a little bit of algebra. Keep this mindset—visualize first, calculate second, and you will conquer any JEE problem!