The locus of z satisfying left|frac{z-i}{z-2i | vedclass

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(B) Given that,$\left|\frac{z-i}{z-2i}\right|=2$. Let $z=x+iy$. Then,$|x+i(y-1)|=2|x+i(y-2)|$. Squaring both sides,we get $x^2+(y-1)^2=4[x^2+(y-2)^2]$

Vedclass · Accepted Answer

Circle

Vedclass · Answer

(B) Given that,$\left|\frac{z-i}{z-2i}\right|=2$. Let $z=x+iy$. Then,$|x+i(y-1)|=2|x+i(y-2)|$. Squaring both sides,we get $x^2+(y-1)^2=4[x^2+(y-2)^2]$. Expanding the terms,$x^2+y^2-2y+1=4[x^2+y^2-4y+4]$. $x^2+y^2-2y+1=4x^2+4y^2-16y+16$. Rearranging the terms,$3x^2+3y^2-14y+15=0$. Dividing by $3$,$x^2+y^2-\frac{14}{3}y+5=0$. This is the equation of a circle in the form $x^2+y^2+2gx+2fy+c=0$. Thus,the locus of $z$ is a circle. Hence,option $(B)$ is correct.

The locus of $z$ satisfying $\left|\frac{z-i}{z-2i}\right|=2$ is a

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