Let z=x+iy represent a point P(x, y) in the A | vedclass

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

Question

(C) The condition $	ext{arg}\left(\frac{z-3}{z-2i}ight)=-\frac{\pi}{2}$ implies that the angle subtended by the segment joining $A(3,0)$ and $B(0,2

Vedclass · Accepted Answer

the arc of the circle $x^2+y^2-3x-2y=0$ intercepted by the diameter $2x+3y-6=0$ not containing the origin and excluding the points $(3,0)$ and $(0,2)$

Vedclass · Answer

(C) The condition $	ext{arg}\left(\frac{z-3}{z-2i}ight)=-\frac{\pi}{2}$ implies that the angle subtended by the segment joining $A(3,0)$ and $B(0,2)$ at point $P(x,y)$ is $-\frac{\pi}{2}$.This means $P$ lies on an arc of a circle passing through $A(3,0)$ and $B(0,2)$.Let $z=x+iy$. Then $\frac{z-3}{z-2i} = \frac{(x-3)+iy}{x+i(y-2)} = \frac{((x-3)+iy)(x-i(y-2))}{x^2+(y-2)^2} = \frac{x(x-3)+y(y-2) + i(xy-2x-3y+6-xy+3y)}{x^2+(y-2)^2} = \frac{x^2+y^2-3x-2y + i(6-2x-3y)}{x^2+(y-2)^2}$.For the argument to be $-\frac{\pi}{2}$,the real part must be $0$ and the imaginary part must be negative.Real part: $x^2+y^2-3x-2y=0$,which is a circle passing through $(3,0)$ and $(0,2)$.Imaginary part: $6-2x-3y  6$.The points $(3,0)$ and $(0,2)$ are excluded because the argument is undefined there.Thus,the locus is the arc of the circle $x^2+y^2-3x-2y=0$ where $2x+3y > 6$.

Let $z=x+iy$ represent a point $P(x, y)$ in the Argand plane. If $z$ satisfies the condition that $\text{arg}\left(\frac{z-3}{z-2i}\right)=-\frac{\pi}{2}$,then the locus of $P$ is

Similar Questions

Let $S = \{z \in \mathbb{C} : |z-1|=1 \text{ and } (\sqrt{2}-1)(z+\bar{z}) - i(z-\bar{z}) = 2\sqrt{2}\}$. Let $z_1, z_2 \in S$ be such that $|z_1| = \max_{z \in S} |z|$ and $|z_2| = \min_{z \in S} |z|$. Then $|\sqrt{2}z_1 - z_2|^2$ equals:

If $|z-2+i| \leq 2$,then the difference between the greatest and least value of $|z|$ is $(i=\sqrt{-1})$.

The minimum value of $|2z - 1| + |3z - 2|$ is

The number of solutions for the equations $|z - 1| = |z - 2| = |z - i|$ is

Let $z$ be a complex number satisfying $|z+5| \leq 4$ and $z(1+i)+\bar{z}(1-i) \geq -10$,where $i=\sqrt{-1}$. If the maximum value of $|z+1|^2$ is $\alpha+\beta \sqrt{2}$,then the value of $(\alpha+\beta)$ is ......

Vedclass Test Series

Exam Paper Generator

Online Exam Module