If z=x+iy and the point P in the Argand plane | vedclass

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(C) Given $z=x+iy$. The equation is $|z-2|+|z-2i|=4$. Substituting $z=x+iy$,we get $|(x-2)+iy|+|x+(y-2)i|=4$. This represents $\sqrt{(x-2)^2+y^2} + \s

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$3x^2+2xy+3y^2-8x-8y=0$

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(C) Given $z=x+iy$. The equation is $|z-2|+|z-2i|=4$. Substituting $z=x+iy$,we get $|(x-2)+iy|+|x+(y-2)i|=4$. This represents $\sqrt{(x-2)^2+y^2} + \sqrt{x^2+(y-2)^2} = 4$. Squaring both sides: $(x-2)^2+y^2 = 16 + x^2+(y-2)^2 - 8\sqrt{x^2+(y-2)^2}$. Simplifying: $x^2-4x+4+y^2 = 16+x^2+y^2-4y+4 - 8\sqrt{x^2+(y-2)^2}$. $-4x+4y-16 = -8\sqrt{x^2+(y-2)^2}$. Dividing by $-4$: $x-y+4 = 2\sqrt{x^2+(y-2)^2}$. Squaring again: $(x-y+4)^2 = 4(x^2+y^2-4y+4)$. $x^2+y^2+16-2xy+8x-8y = 4x^2+4y^2-16y+16$. Rearranging terms: $3x^2+3y^2+2xy-8x-8y=0$.

If $z=x+iy$ and the point $P$ in the Argand plane represents $z$,then the locus of $z$ satisfying the equation $|z-2|+|z-2i|=4$ is

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