The set of all real values of c for which the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The general equation of a circle in the complex plane is given by $z \bar{z} + a \bar{z} + \bar{a} z + b = 0$,where $a$ is a complex constant and

Vedclass · Accepted Answer

$(-\infty, 25]$

Vedclass · Answer

(D) The general equation of a circle in the complex plane is given by $z \bar{z} + a \bar{z} + \bar{a} z + b = 0$,where $a$ is a complex constant and $b$ is a real constant. The centre of this circle is $-a$ and the radius is $\sqrt{|a|^2 - b}$. Comparing the given equation $z \bar{z} + (4 - 3i) \bar{z} + (4 + 3i) z + c = 0$ with the general form,we have $a = 4 - 3i$ and $b = c$. For the equation to represent a circle,the radius must be a real number greater than $0$,so $|a|^2 - b > 0$. Here,$|a|^2 = |4 - 3i|^2 = 4^2 + (-3)^2 = 16 + 9 = 25$. Thus,the condition for the radius is $25 - c > 0$,which implies $c If the radius is allowed to be $0$ (a point circle),then $25 - c \geq 0$,which implies $c \leq 25$. Therefore,the set of all real values of $c$ is $(-\infty, 25]$.

The set of all real values of $c$ for which the equation $z \bar{z} + (4 - 3i) \bar{z} + (4 + 3i) z + c = 0$ represents a circle is

Similar Questions

The locus of the points $z$ which satisfy the condition $\text{arg} \left( \frac{z - 1}{z + 1} \right) = \frac{\pi}{3}$ is

If $z = \frac{3}{2 + \cos \theta + i \sin \theta}$,then the locus of $z$ is :-

If $\sin A+\sin B+\sin C=0$ and $\cos A+\cos B+\cos C=0$,then $\cos (A+B)+\cos (B+C)+\cos (C+A)$ is equal to

If $z$ is a complex number such that $z^2 = (\bar{z})^2$,then

If complex numbers ${z_1}, {z_2}, \text{and } {z_3}$ represent the vertices $A, B, \text{and } C$ respectively of an isosceles triangle $ABC$ of which $\angle C$ is a right angle,then the correct statement is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module