Under what condition do points z = x+iy lie inside a circle…

Under what condition do points $z = x+iy$ lie inside a circle of radius $r$ with center at point $c = a+bi$?

Step 1 · Set up the circle

Center: $c = a+bi$, i.e. the point $(a,b)$. Radius: $r > 0$. A general point: $z = x+iy$, i.e. the point $(x,y)$.

Step 2 · Distance in the complex plane

The distance between $z$ and $c$ is:

$$ \text{distance} = |z - c| $$

Step 3 · Condition for the interior

A point $z$ lies inside the circle (interior, excluding the boundary) exactly when its distance to the center is less than the radius:

$$ \boxed{|z - c| < r} $$

Step 4 · Explicit (Cartesian) form

Substituting $z = x+iy$ and $c = a+bi$:

$$ |z-c| = |(x-a) + i(y-b)| = \sqrt{(x-a)^2 + (y-b)^2} $$

so the condition becomes:

$$ \sqrt{(x-a)^2 + (y-b)^2} < r $$

Step 5 · Equivalent forms

Complex form: $$ \boxed{|z-c| < r} $$

Squared form (squaring is valid since both sides are non-negative): $$ \boxed{|z-c|^2 < r^2 \iff (x-a)^2+(y-b)^2 < r^2} $$

Cartesian (radical) form: $$ \boxed{\sqrt{(x-a)^2+(y-b)^2} < r} $$

Step 6 · Related regions

Step 7 · Worked example

Find the condition for points inside a circle of radius $3$ centered at $c = 2+i$ (so $a=2$, $b=1$, $r=3$):

$$ |z-(2+i)| < 3 \qquad\Longleftrightarrow\qquad (x-2)^2+(y-1)^2 < 9 $$

Step 8 · Verification

Is $z = 2+2i$ inside this circle?

$$ |z-c| = |(2+2i)-(2+i)| = |i| = 1 $$

Since $1 < 3$, yes — $z=2+2i$ is inside. ✓

Is $z = 5+i$ inside?

$$ |z-c| = |(5+i)-(2+i)| = |3| = 3 $$

Since $3 = 3$ (not $<3$), $z=5+i$ lies on the boundary, not inside. ✗