Locus (Complex Numbers)

Question

Question: Let $u = 3 - i$ and $v = 2 + i$ and consider the locus $|z - v| = 2$. Find the minimum value of $|z - u|$ where $z$ is any point on the locus.

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What I have done;

First of all by using

$$ |z-v| = 2 $$

I got a circle centered at $(2 , i)$ with a radius $2$

and by using

$$ |z-u| $$

I have a point located at $(3 , -1)$

After this I have found a straight line that passes through these 2 points with the equation

$$ y = -2x + 5 $$

This line intersects the circle at two points

Shown here

So how do I find the 'minimum' value of $ |z-u| $ , I have to show most things algebraically (as a side note..)

Answer 1

Intuitively, we are trying to find the circle centred at $u$ with the smallest radius such that it is barely touching the circle of radius $2$ centred at $v$. This can be obtained by measuring the (shortest) straight-line distance between $u$ and $v$, and subtracting $2$.

More rigorously, we can use the triangle inequality: $$ |u - v| = |(u - z) + (z - v)| \leq |u - z| + |z - v| $$ Hence, the minimum radius is: $$ |z - u| \geq |u - v| - 2 = \sqrt{(2 - 3)^2 + (1 + 1)^2} - 2 = \sqrt 5 - 2 $$

Answer 2

The circles are centred at $(2,1),(3,-1)$ so the shortest distance the radius of circle formed between these two circles . So its simply subtracting the distance between these two points from $2$ so its $2-\sqrt{(2-3)^2+(1+1)^2}=2-\sqrt{5}$