If z_{1}=2+3i and z_{2}=3+4i are two points o | vedclass

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(C) Given,$z_{1}=2+3i$ and $z_{2}=3+4i$.We have the equation $|z-z_{1}|^{2}+|z-z_{2}|^{2}=|z_{1}-z_{2}|^{2}$.Let $z=x+iy$.Then $|(x-2)+i(y-3)|^{2}+|(x

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a circle

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(C) Given,$z_{1}=2+3i$ and $z_{2}=3+4i$.We have the equation $|z-z_{1}|^{2}+|z-z_{2}|^{2}=|z_{1}-z_{2}|^{2}$.Let $z=x+iy$.Then $|(x-2)+i(y-3)|^{2}+|(x-3)+i(y-4)|^{2}=|(2-3)+i(3-4)|^{2}$.$(x-2)^{2}+(y-3)^{2}+(x-3)^{2}+(y-4)^{2}=|-1-i|^{2}$.$(x^{2}-4x+4)+(y^{2}-6y+9)+(x^{2}-6x+9)+(y^{2}-8y+16)=1+1$.$2x^{2}+2y^{2}-10x-14y+38=2$.$2x^{2}+2y^{2}-10x-14y+36=0$.$x^{2}+y^{2}-5x-7y+18=0$.This is the equation of a circle with center $(\frac{5}{2}, \frac{7}{2})$ and radius $\sqrt{(\frac{5}{2})^{2}+(\frac{7}{2})^{2}-18} = \sqrt{\frac{25+49-72}{4}} = \sqrt{\frac{2}{4}} = \frac{1}{\sqrt{2}}$.

If $z_{1}=2+3i$ and $z_{2}=3+4i$ are two points on the complex plane,then the set of complex numbers $z$ satisfying $|z-z_{1}|^{2}+|z-z_{2}|^{2}=|z_{1}-z_{2}|^{2}$ represents:

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