find center and radius of circle represented by |Z-1| =2 , where z is complex number . About the author Adalyn
Step-by-step explanation: Squaring the given relation, we have (z−α)( z−α )=k 2 (z−β)( z−β ) or (z−α)( z − α )=k 2 (z−β)( z − β ) or z z − α z−α z +α α =k 2 (z z − β z−β z +β β ) or z z (1−k 2 )−( α −k 2 β )z−(α−k 2 β) z +α α −k 2 β β =0 or z z − 1−k 2 α −k 2 β z− 1−k 2 (α−k 2 β) z + 1−k 2 ∣α∣ 2 −k 2 ∣β∣ 2 =0 or z z − a z−a z +a a −a a + 1−k 2 ∣α∣ 2 −k 2 ∣β∣ 2 =0 or ∣z−a∣ 2 =∣a∣ 2 − 1−k 2 ∣α∣ 2 −k 2 ∣ ∣ ∣ β 2 ∣ ∣ ∣ =r 2 Above represents a circle with center at a= 1−k 2 α−k 2 β and r 2 = as written above. Reply
Step-by-step explanation:
Squaring the given relation, we have
(z−α)(
z−α
)=k
2
(z−β)(
z−β
)
or (z−α)(
z
−
α
)=k
2
(z−β)(
z
−
β
)
or z
z
−
α
z−α
z
+α
α
=k
2
(z
z
−
β
z−β
z
+β
β
)
or z
z
(1−k
2
)−(
α
−k
2
β
)z−(α−k
2
β)
z
+α
α
−k
2
β
β
=0
or z
z
−
1−k
2
α
−k
2
β
z−
1−k
2
(α−k
2
β)
z
+
1−k
2
∣α∣
2
−k
2
∣β∣
2
=0
or z
z
−
a
z−a
z
+a
a
−a
a
+
1−k
2
∣α∣
2
−k
2
∣β∣
2
=0
or ∣z−a∣
2
=∣a∣
2
−
1−k
2
∣α∣
2
−k
2
∣
∣
∣
β
2
∣
∣
∣
=r
2
Above represents a circle with center at a=
1−k
2
α−k
2
β
and r
2
= as written above.