(A) The equation of a circle with centre $z_0$ and radius $r$ is given by $|z - z_0| = r$.Squaring both sides,we get $|z - z_0|^2 = r^2$.Using the pro

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$z\bar{z} - z\bar{z_0} - \bar{z}z_0 + z_0\bar{z_0} = r^2$

(A) The equation of a circle with centre $z_0$ and radius $r$ is given by $|z - z_0| = r$.Squaring both sides,we get $|z - z_0|^2 = r^2$.Using the property $|w|^2 = w\bar{w}$,we have $(z - z_0)(\overline{z - z_0}) = r^2$.$(z - z_0)(\bar{z} - \bar{z_0}) = r^2$.Expanding the product,we get $z\bar{z} - z\bar{z_0} - \bar{z}z_0 + z_0\bar{z_0} = r^2$.

Class 11 Mathematics 1 Complex Numbers and Quadratic Equations: Complex Numbers Geometry of complex numbers

Medium

$A$ circle whose radius is $r$ and centre is $z_0$,then the equation of the circle is

A
$z\bar{z} - z\bar{z_0} - \bar{z}z_0 + z_0\bar{z_0} = r^2$
B
$z\bar{z} + z\bar{z_0} - \bar{z}z_0 + z_0\bar{z_0} = r^2$
C
$z\bar{z} - z\bar{z_0} + \bar{z}z_0 - z_0\bar{z_0} = r^2$
D
None of these

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1 Complex Numbers and Quadratic Equations: Complex Numbers

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Geometry of complex numbers

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Let $C$ be the set of all complex numbers. Let $S_{1} = \{z \in C : |z-3-2i|^{2}=8\}$,$S_{2} = \{z \in C : \operatorname{Re}(z) \geq 5\}$,and $S_{3} = \{z \in C : |z-\bar{z}| \geq 8\}$. Then the number of elements in $S_{1} \cap S_{2} \cap S_{3}$ is equal to:

DifficultJEE MAIN 2021

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For any two complex numbers $z_1, z_2$,if $|z_1 + z_2|^2 = |z_1|^2 + |z_2|^2$,then:

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$A$ circle whose radius is $r$ and centre is $z_0$,then the equation of the circle is

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The equation of the locus of $z$ such that $\left|\frac{z-i}{z+i}\right|=2$,where $z=x+iy$ is a complex number,is

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