
The equation $\left| {z - {z_ \circ }} \right| = r$ represents
(Given: ${z_ \circ }$ is a fixed complex number)
A.A line
B.A circle with centre z0 and radius r
C.A circle with centre (0,0) and radius 1
D.A line through origin
Answer
545.1k+ views
Hint: Here we need to determine whether the given equation is of line or circle. We will first assume the fixed complex number which is present in the equation and the variable complex number of the equation. Then we will substitute these values in the equation and then we will solve the equation and from there, we will get the required answer.
Complete step-by-step answer:
The given equation is $\left| {z - {z_ \circ }} \right| = r$. We need to find the nature of $z$.
Here it is given that ${z_ \circ }$ is a fixed complex number, which means that this complex number represents only a single point on the complex plane.
Here $z$ is a variable complex number.
Let ${z_ \circ } = {x_ \circ } + i{y_ \circ }$ , where ${x_ \circ }$ and ${y_ \circ }$ are constants and
$z = x + iy$, where $x$ and $y$ are constants.
Now, we will substitute these values in the given equation.
$ \Rightarrow \left| {x + iy - {x_ \circ } - i{y_ \circ }} \right| = r$
Subtracting real part from real part and imaginary part from imaginary part, we get
$ \Rightarrow \left| {\left( {x - {x_ \circ }} \right) + i\left( {y - {y_ \circ }} \right)} \right| = r$
On simplifying the modulus of this complex number, we get
$ \Rightarrow \sqrt {{{\left( {x - {x_ \circ }} \right)}^2} + {{\left( {y - {y_ \circ }} \right)}^2}} = r$
On squaring both sides, we get
$ \Rightarrow {\left( {x - {x_ \circ }} \right)^2} + {\left( {y - {y_ \circ }} \right)^2} = {r^2}$
We know that this equation represents the equation of a circle with center as $\left( {{x_ \circ },{y_ \circ }} \right)$ and the radius equal to $r$
Since, the point $\left( {{x_ \circ },{y_ \circ }} \right)$ is represented by the complex number ${z_ \circ }$, so we can say that the center of the circle is ${z_ \circ }$.
Hence, the correct option is option B.
Note: Since we have used complex numbers here and also we have solved the complex equations, so we need to know its definitions. A complex number is defined as the number which is the combination of both real number and an imaginary number and the complex number is represented by $a + ib$ .
Complete step-by-step answer:
The given equation is $\left| {z - {z_ \circ }} \right| = r$. We need to find the nature of $z$.
Here it is given that ${z_ \circ }$ is a fixed complex number, which means that this complex number represents only a single point on the complex plane.
Here $z$ is a variable complex number.
Let ${z_ \circ } = {x_ \circ } + i{y_ \circ }$ , where ${x_ \circ }$ and ${y_ \circ }$ are constants and
$z = x + iy$, where $x$ and $y$ are constants.
Now, we will substitute these values in the given equation.
$ \Rightarrow \left| {x + iy - {x_ \circ } - i{y_ \circ }} \right| = r$
Subtracting real part from real part and imaginary part from imaginary part, we get
$ \Rightarrow \left| {\left( {x - {x_ \circ }} \right) + i\left( {y - {y_ \circ }} \right)} \right| = r$
On simplifying the modulus of this complex number, we get
$ \Rightarrow \sqrt {{{\left( {x - {x_ \circ }} \right)}^2} + {{\left( {y - {y_ \circ }} \right)}^2}} = r$
On squaring both sides, we get
$ \Rightarrow {\left( {x - {x_ \circ }} \right)^2} + {\left( {y - {y_ \circ }} \right)^2} = {r^2}$
We know that this equation represents the equation of a circle with center as $\left( {{x_ \circ },{y_ \circ }} \right)$ and the radius equal to $r$
Since, the point $\left( {{x_ \circ },{y_ \circ }} \right)$ is represented by the complex number ${z_ \circ }$, so we can say that the center of the circle is ${z_ \circ }$.
Hence, the correct option is option B.
Note: Since we have used complex numbers here and also we have solved the complex equations, so we need to know its definitions. A complex number is defined as the number which is the combination of both real number and an imaginary number and the complex number is represented by $a + ib$ .
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

