Q.Show that the complex number $z$, satisfying the condition $\arg\left(\dfrac{z-1}{z+1}\right)=\dfrac{\pi}{4}$ lies on a circle.

Circles

Tie a stone to a string and swing it around your head: the stone traces a loop where every point sits the same distance from your hand. That is a circle — the set of all points in a plane at a fixed distance from a fixed point.

The fixed point is the centre $O$; the fixed distance is the radius $r$. For any point $P$ on the circle, $OP = r$. In locus language, a circle is the locus of a point that moves so that its distance from the centre stays constant.

The standard equation

Put the centre at the origin and let $P(x,y)$ be any point on the circle. Its distance from the centre is $\sqrt{x^2+y^2}=r$. Squaring both sides:

If the centre sits at $(h,k)$ instead, the distance formula gives the standard form

Every choice of centre and radius produces exactly one such equation, and every point satisfying it lies on the circle.

A quick check

For $x^2+y^2=25$ the centre is $(0,0)$ and $r=5$:

• $(3,4)$: $9+16=25$ ✓ on the circle
• $(1,2)$: $1+4=5\neq 25$ ✗ not on the circle …