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Area of triangle, circle (formula) - (Concept)
Equation of Circle
The equation of the circle whose centre is at the point $z_0$ and has radius $r$ is given by
$
\left|z-z_0\right|=r
$
If the center is origin then, $z_0=0$, hence equation reduces to $|z|=r$
Interior of the circle is represented by $\left|z-z_0\right|<r$
The exterior is represented by $\left|z-z_0\right|>r$
Here z can be represented as $\mathrm{x}+\mathrm{iy}$ and $z_0$ is represented by $x_0+i y_0$
Equation of Circle in second form
$\frac{\left|z-z_1\right|}{\left|z-z_2\right|}=k \quad(k \neq 1, k>0)$
This equation also represents a circle. This can be verified by putting z = x+iy, z1 = p+iq, z2 = a+ib
Equation of Ellipse
$\left|z-z_1\right|+\left|z-z_2\right|=k \quad\left(k>\left|z_1-z_2\right|\right)$
This represents an ellipse as the sum of distances of point z from z1 and z2 is constant, which is the locus of an ellipse.
Equation of Hyperbola
$\left|\left|z-z_1\right|-\left|z-z_2\right|\right|=k \quad\left(k<\left|z_1-z_2\right|\right)$
This represents a hyperbola as the difference of distances of point z from z1 and z2 is constant, which is the locus of a hyperbola.
Section Formula
The complex number z dividing z1 and z2 internally in ratio m: n is given by
$
\mathrm{z}=\frac{m z_2+n z_1}{m+n}
$
And
The complex number $z$ dividing $z_1$ and $z_2$ externally in ratio $m$ : $n$ is given by
$
\mathrm{z}=\frac{m z_2-n z_1}{m-n}
$
Centroid of the triangle with vertices z1, z2 and z3 is given by $\left(z_1+z_2+z_3\right) / 3$
| Exam | Chapter |
| JEE MAIN | Complex numbers and quadratic equations |
Let and
are equations of two circles, then number of common solutions between them equals
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The largest value of r for which the region represented by the set $\{w \epsilon C /|w-4-i| \leq r\}$ is contained in the region represented by the set $\{z \epsilon C /|z-1| \leq|z+i|\}$ is equal to:
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The locus of the centre of a circle which touches the circle $\left|z-z_1\right|=a$ and $\left|z-z_2\right|=b$ externally $\left(z, z_1, z_2\right.$ are complex number $)$ will be
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If $z$ be a complex number satisfying $|\operatorname{Re}(z)|+|\operatorname{Im}(z)|=4$, then $|z| \ {\text {cannot }}$ be :
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Let $z$ be a complex number such that $\left|\frac{z-i}{z+2 i}\right|=1 \quad|z|=\frac{5}{2}$. Then the value of $|z+3 i|_{\text {is : }}$
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If the equation represents a circle where a, d are real constants, then which of the following condition is correct?
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The minimum distance of from
equals $\sqrt{n}-2$, then the value of $n$ is
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Let $z_1$ and $z_2$ be two complex numbers satisfying $\left|z_1\right|=9$ and $\left|z_2-3-4 i\right|=4$. Then the minimum value of $\left|z_1-z_2\right|$ is:
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Let $S_1, S_2$ and $S_3$ be three sets defined as
$
\begin{aligned}
& S_1=\{z \in \mathbb{C}:|z-1| \leq \sqrt{2}\} \\
& S_2=\{z \in \mathbb{C}: \operatorname{Re}((1-i) z) \geqslant 1\} \\
& S_3=\{z \in \mathbb{C}: \operatorname{Im}(z) \leq 1\}
\end{aligned}
$
Then the set $S_1 \cap S_2 \cap S_3$
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$\mathrm{Let\; A= \left \{ z\in C:1\leqslant \left | z-\left ( 1+i \right ) \right |\leqslant 2 \right \} }$ $\mathrm{and\; B= \left \{ z\in A:\left | z-\left ( 1-i \right ) \right |= 1 \right \} }$. Then, B:
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Let $\mathrm{S}=\{\mathrm{z} \in \mathbb{C}:|\mathrm{z}-3| \leq 1$ and $\mathrm{z}(4+3 \mathrm{i})+\overline{\mathrm{z}}(4-3 \mathrm{i}) \leq 24\}$. If $\alpha+\mathrm{i} \beta$ is the point in S which is closest to 4i , then $25(\alpha+\beta)$ is equal to
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Let a circle C in complex plane pass through the points $z_1=3+4 i, z_2=4+3 i$ and $z_3=5 i_{\text {.If }} z\left(\neq z_1\right)$ is a point on C such that the line through $z$ and $z_1$ is perpendicular to the line through $z_2$ and $z_3$, then arg $(\mathrm{z})$ is equal to:
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Let $\mathrm{S}=\{z \in \mathrm{C}:|z-2| \leq 1, z(1+i)+\bar{z}(1-i) \leq 2\}$. Let $|z-4 i|$ attains minimum and maximum values, respectively, at $z_1 \in S$ and $z_2 \in \mathrm{~S}$. If $5\left(\left|z_1\right|^2+\left|z_2\right|^2\right)=\alpha+\beta \sqrt{5}$, where $\alpha$ and $\beta$ are integers, then the value of $\alpha+\beta$ is equal to $\qquad$ .
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For , let
and
. Then, the number of elements in the set
is:
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If $\mathrm{z}=\mathrm{x}+\mathrm{iy}$ satisfies $|\mathrm{z}|-2=0$ and $|\mathrm{z}-\mathrm{i}|-|\mathrm{z}+5 \mathrm{i}|=0$, then
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Let $\mathrm{S}=\{z=x+i y:|z-1+i| \geq|z|,|z|<2,|z+i|=|z-1|\}$. Then the set of all values of x , for which $w=2 \mathrm{x}+\mathrm{iy} \in \mathrm{S}$ for some $\mathrm{y} \in \mathbb{R}$, is
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Let $C$ be the set of all complex numbers. Let
$
\begin{aligned}
& S_1=\left\{z \in C| | z-3-\left.2 i\right|^2=8\right\}, \\
& S_2=\{z \in C \mid \operatorname{Re}(z) \geq 5\} \text { and } \\
& S_3=\{z \in C| | z-\bar{z} \mid \geq 8\} .
\end{aligned}
$
Then the number of elements in $S_1 \cap S_2 \cap S_3$ is equal to
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The equation arg $\left ( \frac{z-1}{z+1} \right )= \frac{\pi }{4}$ represents a circle with
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A point $z$ moves in the complex plane such that $\arg \left(\frac{z-2}{z+2}\right)=\frac{\pi}{4}$, then the minimum value of $|z-9 \sqrt{2}-2 i|^2$ is equal to .
$\qquad$
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If $z$ is a complex number such that $\frac{z-i}{z-1}$ is purely imaginary, then the minimum value of $|z-(3+3 i)|$ is :
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If for the complex numbers $z$ satisfying $|z-2-2 l| \leq 1$, the maximum value of $|3 i z+6|$ is attained at $a+i b$, then $a+b$ is equal to $\qquad$
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If $|z-3+2 i| \leq 4$ then the difference between the greatest value and the least value of $|z|_{\text {is : }}$
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The minimum distance of $\mathrm{z}_1=3+4 i$ from $|\mathrm{z}-1-i|=2$ equals to $\sqrt{n}-2$. Then find the value of $n$.
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Let , the set of complex numbers. Then the equation,
represents :
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If $\operatorname{Re}\left[\frac{z-1}{2 z+i}\right]=1$, where $z=x+i y$, then the point $(x, y)$ lies on a :
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If Re where
then the point
lies on a :
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Let z be a complex number such that . Then z lies on the circle of radius 2 and centre
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$\begin{aligned} & \alpha=8-14 i, A=\left\{z \in \mathbb{C}: \frac{\alpha z-\bar{\alpha} \bar{z}}{z^2-(\bar{z})^2-112 i}=1\right\} \text { and } \\ & B=\{z \in \mathbb{C}:|z+3 i|=4\} \text {. Then } \sum_{z \in A \cap B}(\operatorname{Re} z-\operatorname{Im} z) \text { is equal to }\end{aligned}$
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For all $z ∈ C$ on the curve $C_1: |z| = 4$, let the locus of the point $z+\frac{1}{z}$ be the curve $C_2$. Then :
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If the center and radius of the circle are respectively
and
. then
is equal to
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For and
, if
is the radius of the circle
, then
is equal to
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Let C be the circle in the complex plane with centre and radius
and the complex number
be outside the circle C such that
. If
and
are collinear, then the smaller value of
is equal to:
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|
Let Let
be the circle C of radius 1 in the first quadrant touching the line
and the y-axis. If the curve
intersects C at A and B, then
is equal to_____________
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Let $A$ and $B$ be the sets defined by $A=\{z:|z| \leq 2, z \in C\}$ and $B=\{z:(1-i) z+(1+i) \bar{z} \geqslant 4, z \in C\}$. Find the area of the region $A \cap B$.
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Write the equation of Circle with $(-1,2)$ as its centre and 2 as its radius, in argand plane.
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Smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line x+y=2 is
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Let $P=\{z \in C:|z+2-3 i| \leq 1\}$ and $Q=\{z \in C: z(1+i)+\bar{z}(1-i) \leq-8\}$. Let in $P \cap Q,|z-3+2 i|$ be maximum and minimum at $z_1$ and $z_2$ respectively. If $\left|z_1\right|^2+2\left|z_2\right|^2=\alpha+\beta \sqrt{2}$, where $\alpha, \beta$ are integers, then $\alpha+\beta$ equals_______.
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The area (in sq. units) of the region $\mathrm{S}=\{\mathrm{z} \in \mathbb{C} ;|\mathrm{z}-1| \leq 2 ;(\mathrm{z}+\overline{\mathrm{z}})+\mathrm{i}(\mathrm{z}-\overline{\mathrm{z}}) \leq 2, \operatorname{lm}(\mathrm{z}) \geq 0\}$ is
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Number of complex number $z$ satisfying $|z+3+i|=|z+9+i|$ and $|z+3+3 i|=3$ is/are:
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|
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|
Let the curve $z(1+i)+\bar{z}(1-i)=4, z \in C$, divide the region $|z-3| \leq 1$ into two parts of areas $\alpha$ and $\beta$. Then $|\alpha-\beta|$ equals:
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|
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|
Let $\left|\frac{\overline{\mathrm{z}}-\mathrm{i}}{2 \overline{\mathrm{z}}+\mathrm{i}}\right|=\frac{1}{3}, \mathrm{z} \in \mathbb{C}$, be the equation of a circle with center at C . If the area of the triangle, whose vertices are at the points $(0,0), \mathrm{C}$ and $(\alpha, 0)$ is 11 square units, then $\alpha^2$ equals
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