# Locus (Complex Numbers)

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.

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..)

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: $$|z - u| \geq |u - v| - 2 = \sqrt{(2 - 3)^2 + (1 + 1)^2} - 2 = \sqrt 5 - 2$$

• Would my method work tho? – bigfocalchord Dec 25 '15 at 4:59
• Well you haven't entirely finished your method, so I'm not sure. I don't see how knowing the equation of the line is helpful though; at some point, you'll need to calculate a distance. – Adriano Dec 25 '15 at 5:09
• I think you should check your root calculation may be a typo – Archis Welankar Dec 25 '15 at 6:45

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 $2-\sqrt{(2-3)^2+(1+1)^2}=2-\sqrt{5}$

• By why is it 2 - sqr(5)? I don't see where the 2 comes from – bigfocalchord Dec 25 '15 at 7:01
• Its subtracting the radius of $2,1$ circle as we have found out distance between the centres of two circles. – Archis Welankar Dec 25 '15 at 7:05
• But why do we need to do 2 minus the distance , do we not only need the distance , I can't seem to see this on the graph >_> why does the radius come into play? – bigfocalchord Dec 25 '15 at 7:07
• its counted twice while calculating the length – Archis Welankar Dec 25 '15 at 7:08