28
Question
|Z−Z1|2+|Z−Z2|2=a represents a real circle [with center Z1+Z22 on the Argand plane.The prove that 2a≥|Z1−Z2|2.
The prove that 2a≥|Z1−Z2|2.
Open in App
Solution
|Z−Z1|2+|Z−Z2|2=a
(Z−Z1)(¯¯¯¯Z−¯¯¯¯Z1)+(Z−Z2)(¯¯¯¯Z−¯¯¯¯Z2)=a
2Z¯¯¯¯Z−Z(¯¯¯¯Z1+¯¯¯¯Z2)−¯¯¯¯Z(Z1+Z2)+Z1¯¯¯¯Z1+Z2¯¯¯¯Z2=a
Z¯¯¯¯Z−(¯¯¯¯Z1+¯¯¯¯Z22)Z−(Z1+Z22)¯¯¯¯Z+Z1¯¯¯¯Z1+Z2¯¯¯¯Z2−a2=0 (1)
Equation (1) is of the form of Z¯¯¯¯Z+¯¯¯¯αZ+α¯¯¯¯Z+r=0. Hence, center = -coefficient of ¯¯¯¯Z; which is given by (Z1+Z2)/2. Also, Eq. (1) will represent a real circle if α¯¯¯¯α−r>0.
⟹(Z1+Z2)(¯¯¯¯Z1+¯¯¯¯Z2)4≥Z1¯¯¯¯Z1+Z2¯¯¯z2−a2
Z1¯¯¯¯Z+Z1¯¯¯¯Z2+¯¯¯¯Z1Z2+Z2¯¯¯¯Z2>2(Z1¯¯¯¯Z1+Z2¯¯¯¯Z2)−2a
2a≥Z1¯¯¯¯Z1+Z2¯¯¯¯Z2−Z1¯¯¯¯Z2−Z2¯¯¯¯Z1
=Z1(¯¯¯¯Z1−¯¯¯¯Z2)−Z2(¯¯¯¯Z1−¯¯¯¯Z2)=(Z1−Z2)(¯¯¯¯Z1−¯¯¯¯Z2)=(Z1−Z2)(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯Z1−Z2)
2a≥|Z1−Z2|2
Ans: 1
(Z−Z1)(¯¯¯¯Z−¯¯¯¯Z1)+(Z−Z2)(¯¯¯¯Z−¯¯¯¯Z2)=a
2Z¯¯¯¯Z−Z(¯¯¯¯Z1+¯¯¯¯Z2)−¯¯¯¯Z(Z1+Z2)+Z1¯¯¯¯Z1+Z2¯¯¯¯Z2=a
Z¯¯¯¯Z−(¯¯¯¯Z1+¯¯¯¯Z22)Z−(Z1+Z22)¯¯¯¯Z+Z1¯¯¯¯Z1+Z2¯¯¯¯Z2−a2=0 (1)
Equation (1) is of the form of Z¯¯¯¯Z+¯¯¯¯αZ+α¯¯¯¯Z+r=0. Hence, center = -coefficient of ¯¯¯¯Z; which is given by (Z1+Z2)/2. Also, Eq. (1) will represent a real circle if α¯¯¯¯α−r>0.
⟹(Z1+Z2)(¯¯¯¯Z1+¯¯¯¯Z2)4≥Z1¯¯¯¯Z1+Z2¯¯¯z2−a2
Z1¯¯¯¯Z+Z1¯¯¯¯Z2+¯¯¯¯Z1Z2+Z2¯¯¯¯Z2>2(Z1¯¯¯¯Z1+Z2¯¯¯¯Z2)−2a
2a≥Z1¯¯¯¯Z1+Z2¯¯¯¯Z2−Z1¯¯¯¯Z2−Z2¯¯¯¯Z1
=Z1(¯¯¯¯Z1−¯¯¯¯Z2)−Z2(¯¯¯¯Z1−¯¯¯¯Z2)=(Z1−Z2)(¯¯¯¯Z1−¯¯¯¯Z2)=(Z1−Z2)(¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯Z1−Z2)
2a≥|Z1−Z2|2
Ans: 1
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
