Given,
z0=21+3i,z1=(1+i)
So, ∣z1−z0∣=(1−21)2+(1−23)2=41+41=21
And z2 be outside circle, ∣z1−z0∣∣z2−z0∣=1
⇒21∣z2−z0∣=1
⇒∣z2−z0∣=2
Now given, {z}_{0},{z}_{1}&{z}_{2} are collinear,
So, by concept of rotation we get,
z1−z0z2−z0=∣z1−z0∣∣z2−z0∣ei0
⇒z1−z0z2−z0=∣z1−z0∣∣z2−z0∣(±1)
⇒z1−z0z2−z0=±2
⇒z2=z0±2(z1−z0)
So, z2=2z1−z0=23+21i⇒∣z2∣2=25
Or z2=3z0−2z1=2−1+25i⇒∣z2∣2=213
Hence, the smaller value will be, ∣z2∣2=25