NEET Test Series from KOTA - 10 Papers In MS WORD
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Complex Numbers and Quadratic Equation
117976
In the argand diagram, all the complex numbers \(z\) satisfying \(|z-4 i|+|z+4 i|=10\) lie on a/an
1 straight line
2 circle
3 ellipse
4 parabola
Explanation:
C Given, \(|z-4 i|+|z+4 i|=10\) \(\text { Put, } z=x+i y\) \(\therefore |x+i y-4 i|+|x+i y+4 i|=10\) \(\text { or } |x+i(y-4)|+|x+i(y+4)|=10\) \(\sqrt{x^2+(y-4)^2}+\sqrt{x^2+(y+4)^2}=10\) \(\text { or } \sqrt{x^2+(y-4)^2}=10-\sqrt{x^2+(y+4)^2}\) \(\text { or } x^2+(y-4)^2=100+x^2+(y+4)^2\) \(\quad-20 \cdot \sqrt{x^2+(y+4)^2}\) \(\text { or } 100+x^2+y^2+16+8 y-x^2-y^2-16+8 y\) \(=20 \cdot \sqrt{x^2+(y+4)^2}\) \((100+16 y)=20 \cdot \sqrt{x^2+(y+4)^2}\) \((25+4 y)=5 \sqrt{x^2+(y+4)^2}\) Squaring again both the sides, we get \((25+4 y)^2=25\left(x^2+(y+4)^2\right)\) \((25+4 y)^2=25\left(x^2+\left(y^2+16+8 y\right)\right)\) \((25)^2+16 y^2+200 y=25 x^2+25 y^2+25 \times 16\) \(+25 \times 8 \mathrm{y}\) \(\therefore \quad 625-25 \times 16=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25(25-16)=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25 \times 9=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1\) which is an ellips.
EAMCET-1996
Complex Numbers and Quadratic Equation
117977
\(P\) represents \(z=x+\) iy in argand plane and \(|3 z-1|=3|z-2|\), then locus of \(P\) is
1 \(x=0\)
2 \(x^2+y^2=8\)
3 \(y=x\)
4 \(6 x=7\)
Explanation:
D \(\mathrm{P}\) represents \(\mathrm{z}=\mathrm{x}+\) iy in the argand plane. And, \(z=x+i y\), Now, \(|3 z-1|=3|z-2|\) \(|3 \mathrm{x}+3 \mathrm{iy}-1|=3|\mathrm{x}+\mathrm{iy}-2|\) or \(\quad|(3 \mathrm{x}-1)+3 \mathrm{y} \cdot \mathrm{i}|=3|(\mathrm{x}-2)+\mathrm{iy}|\) or \(\quad(3 \mathrm{x}-1)^2+(3 \mathrm{y})^2=9\left[(\mathrm{x}-2)^2+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9\left[\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9 \mathrm{x}^2+9 \mathrm{y}^2+36-36 \mathrm{x}\) \(30 \mathrm{x}=35\) or \(\mathrm{x}=\frac{35}{30}=\frac{7}{6}\) \(6 \mathrm{x}=7\)
EAMCET-1999
Complex Numbers and Quadratic Equation
117978
The locus of the \(\mathrm{z}\) in the argand plane for which \(|z+1|^2+|z-1|^2=4\), is a
1 straight line
2 pair of straight lines
3 circle
4 parabola
Explanation:
C Given, \(|\mathrm{z}+1|^2+|\mathrm{z}-1|^2=4\) \(\text { Put, } \mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\therefore |(\mathrm{x}+1)+\mathrm{iy}|^2+|(\mathrm{x}-1)+\mathrm{iy}|^2=4\) \(\text { or } (\mathrm{x}+1)^2+\mathrm{y}^2+(\mathrm{x}-1)^2+\mathrm{y}^2=4\) \(\text { or } \mathrm{x}^2+1+2 \mathrm{x}+\mathrm{y}^2+\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=4\) \(\text { or } 2 \mathrm{x}^2+2 \mathrm{y}^2+2=4\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2+1=2\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2=1 \text { which is a circle. }\)
EAMCET-2000
Complex Numbers and Quadratic Equation
117979
If \(z=x+i y, x, y \in[R\) and if the point \(P\) in the Argand plane represents \(z\), then the locus of \(P\) satisfying the condition \(\operatorname{Arg}\left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) is
1 \(\left\{z \in C /\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
2 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6=0\}\)
3 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6>0\), \(\left.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
4 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6\lt 0\), \(\left.\left|z-\frac{1+3 \mathrm{i}}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
Explanation:
C We have, \(\arg \left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) \(\arg (z-1)-\arg (z-3 i)=\frac{\pi}{2}\) \(\arg [(x-1)+i y]-\arg [x+(y-3) i]=\frac{\pi}{2}\) \(\tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y-3}{x}=\frac{\pi}{2}\) \(\tan ^{-1}\left[\frac{\mathrm{y}-1}{1+\frac{y}{x} \times \frac{y}{x}}\right]=\frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(y-3)}=\frac{1}{0}\) \(x(x-1)+y(y-3)=0\) \(x^2+y^2-x-3 y=0\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{1}{4}+\frac{9}{4}\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{\sqrt{10}}{2}\right)^2\) Which is circle with centre \(\left(\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{\sqrt{10}}{2}\) \(\therefore z \in C:(3-i) z+(3+i) \bar{z}-6>0,\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\)
117976
In the argand diagram, all the complex numbers \(z\) satisfying \(|z-4 i|+|z+4 i|=10\) lie on a/an
1 straight line
2 circle
3 ellipse
4 parabola
Explanation:
C Given, \(|z-4 i|+|z+4 i|=10\) \(\text { Put, } z=x+i y\) \(\therefore |x+i y-4 i|+|x+i y+4 i|=10\) \(\text { or } |x+i(y-4)|+|x+i(y+4)|=10\) \(\sqrt{x^2+(y-4)^2}+\sqrt{x^2+(y+4)^2}=10\) \(\text { or } \sqrt{x^2+(y-4)^2}=10-\sqrt{x^2+(y+4)^2}\) \(\text { or } x^2+(y-4)^2=100+x^2+(y+4)^2\) \(\quad-20 \cdot \sqrt{x^2+(y+4)^2}\) \(\text { or } 100+x^2+y^2+16+8 y-x^2-y^2-16+8 y\) \(=20 \cdot \sqrt{x^2+(y+4)^2}\) \((100+16 y)=20 \cdot \sqrt{x^2+(y+4)^2}\) \((25+4 y)=5 \sqrt{x^2+(y+4)^2}\) Squaring again both the sides, we get \((25+4 y)^2=25\left(x^2+(y+4)^2\right)\) \((25+4 y)^2=25\left(x^2+\left(y^2+16+8 y\right)\right)\) \((25)^2+16 y^2+200 y=25 x^2+25 y^2+25 \times 16\) \(+25 \times 8 \mathrm{y}\) \(\therefore \quad 625-25 \times 16=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25(25-16)=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25 \times 9=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1\) which is an ellips.
EAMCET-1996
Complex Numbers and Quadratic Equation
117977
\(P\) represents \(z=x+\) iy in argand plane and \(|3 z-1|=3|z-2|\), then locus of \(P\) is
1 \(x=0\)
2 \(x^2+y^2=8\)
3 \(y=x\)
4 \(6 x=7\)
Explanation:
D \(\mathrm{P}\) represents \(\mathrm{z}=\mathrm{x}+\) iy in the argand plane. And, \(z=x+i y\), Now, \(|3 z-1|=3|z-2|\) \(|3 \mathrm{x}+3 \mathrm{iy}-1|=3|\mathrm{x}+\mathrm{iy}-2|\) or \(\quad|(3 \mathrm{x}-1)+3 \mathrm{y} \cdot \mathrm{i}|=3|(\mathrm{x}-2)+\mathrm{iy}|\) or \(\quad(3 \mathrm{x}-1)^2+(3 \mathrm{y})^2=9\left[(\mathrm{x}-2)^2+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9\left[\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9 \mathrm{x}^2+9 \mathrm{y}^2+36-36 \mathrm{x}\) \(30 \mathrm{x}=35\) or \(\mathrm{x}=\frac{35}{30}=\frac{7}{6}\) \(6 \mathrm{x}=7\)
EAMCET-1999
Complex Numbers and Quadratic Equation
117978
The locus of the \(\mathrm{z}\) in the argand plane for which \(|z+1|^2+|z-1|^2=4\), is a
1 straight line
2 pair of straight lines
3 circle
4 parabola
Explanation:
C Given, \(|\mathrm{z}+1|^2+|\mathrm{z}-1|^2=4\) \(\text { Put, } \mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\therefore |(\mathrm{x}+1)+\mathrm{iy}|^2+|(\mathrm{x}-1)+\mathrm{iy}|^2=4\) \(\text { or } (\mathrm{x}+1)^2+\mathrm{y}^2+(\mathrm{x}-1)^2+\mathrm{y}^2=4\) \(\text { or } \mathrm{x}^2+1+2 \mathrm{x}+\mathrm{y}^2+\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=4\) \(\text { or } 2 \mathrm{x}^2+2 \mathrm{y}^2+2=4\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2+1=2\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2=1 \text { which is a circle. }\)
EAMCET-2000
Complex Numbers and Quadratic Equation
117979
If \(z=x+i y, x, y \in[R\) and if the point \(P\) in the Argand plane represents \(z\), then the locus of \(P\) satisfying the condition \(\operatorname{Arg}\left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) is
1 \(\left\{z \in C /\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
2 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6=0\}\)
3 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6>0\), \(\left.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
4 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6\lt 0\), \(\left.\left|z-\frac{1+3 \mathrm{i}}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
Explanation:
C We have, \(\arg \left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) \(\arg (z-1)-\arg (z-3 i)=\frac{\pi}{2}\) \(\arg [(x-1)+i y]-\arg [x+(y-3) i]=\frac{\pi}{2}\) \(\tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y-3}{x}=\frac{\pi}{2}\) \(\tan ^{-1}\left[\frac{\mathrm{y}-1}{1+\frac{y}{x} \times \frac{y}{x}}\right]=\frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(y-3)}=\frac{1}{0}\) \(x(x-1)+y(y-3)=0\) \(x^2+y^2-x-3 y=0\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{1}{4}+\frac{9}{4}\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{\sqrt{10}}{2}\right)^2\) Which is circle with centre \(\left(\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{\sqrt{10}}{2}\) \(\therefore z \in C:(3-i) z+(3+i) \bar{z}-6>0,\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\)
117976
In the argand diagram, all the complex numbers \(z\) satisfying \(|z-4 i|+|z+4 i|=10\) lie on a/an
1 straight line
2 circle
3 ellipse
4 parabola
Explanation:
C Given, \(|z-4 i|+|z+4 i|=10\) \(\text { Put, } z=x+i y\) \(\therefore |x+i y-4 i|+|x+i y+4 i|=10\) \(\text { or } |x+i(y-4)|+|x+i(y+4)|=10\) \(\sqrt{x^2+(y-4)^2}+\sqrt{x^2+(y+4)^2}=10\) \(\text { or } \sqrt{x^2+(y-4)^2}=10-\sqrt{x^2+(y+4)^2}\) \(\text { or } x^2+(y-4)^2=100+x^2+(y+4)^2\) \(\quad-20 \cdot \sqrt{x^2+(y+4)^2}\) \(\text { or } 100+x^2+y^2+16+8 y-x^2-y^2-16+8 y\) \(=20 \cdot \sqrt{x^2+(y+4)^2}\) \((100+16 y)=20 \cdot \sqrt{x^2+(y+4)^2}\) \((25+4 y)=5 \sqrt{x^2+(y+4)^2}\) Squaring again both the sides, we get \((25+4 y)^2=25\left(x^2+(y+4)^2\right)\) \((25+4 y)^2=25\left(x^2+\left(y^2+16+8 y\right)\right)\) \((25)^2+16 y^2+200 y=25 x^2+25 y^2+25 \times 16\) \(+25 \times 8 \mathrm{y}\) \(\therefore \quad 625-25 \times 16=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25(25-16)=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25 \times 9=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1\) which is an ellips.
EAMCET-1996
Complex Numbers and Quadratic Equation
117977
\(P\) represents \(z=x+\) iy in argand plane and \(|3 z-1|=3|z-2|\), then locus of \(P\) is
1 \(x=0\)
2 \(x^2+y^2=8\)
3 \(y=x\)
4 \(6 x=7\)
Explanation:
D \(\mathrm{P}\) represents \(\mathrm{z}=\mathrm{x}+\) iy in the argand plane. And, \(z=x+i y\), Now, \(|3 z-1|=3|z-2|\) \(|3 \mathrm{x}+3 \mathrm{iy}-1|=3|\mathrm{x}+\mathrm{iy}-2|\) or \(\quad|(3 \mathrm{x}-1)+3 \mathrm{y} \cdot \mathrm{i}|=3|(\mathrm{x}-2)+\mathrm{iy}|\) or \(\quad(3 \mathrm{x}-1)^2+(3 \mathrm{y})^2=9\left[(\mathrm{x}-2)^2+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9\left[\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9 \mathrm{x}^2+9 \mathrm{y}^2+36-36 \mathrm{x}\) \(30 \mathrm{x}=35\) or \(\mathrm{x}=\frac{35}{30}=\frac{7}{6}\) \(6 \mathrm{x}=7\)
EAMCET-1999
Complex Numbers and Quadratic Equation
117978
The locus of the \(\mathrm{z}\) in the argand plane for which \(|z+1|^2+|z-1|^2=4\), is a
1 straight line
2 pair of straight lines
3 circle
4 parabola
Explanation:
C Given, \(|\mathrm{z}+1|^2+|\mathrm{z}-1|^2=4\) \(\text { Put, } \mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\therefore |(\mathrm{x}+1)+\mathrm{iy}|^2+|(\mathrm{x}-1)+\mathrm{iy}|^2=4\) \(\text { or } (\mathrm{x}+1)^2+\mathrm{y}^2+(\mathrm{x}-1)^2+\mathrm{y}^2=4\) \(\text { or } \mathrm{x}^2+1+2 \mathrm{x}+\mathrm{y}^2+\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=4\) \(\text { or } 2 \mathrm{x}^2+2 \mathrm{y}^2+2=4\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2+1=2\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2=1 \text { which is a circle. }\)
EAMCET-2000
Complex Numbers and Quadratic Equation
117979
If \(z=x+i y, x, y \in[R\) and if the point \(P\) in the Argand plane represents \(z\), then the locus of \(P\) satisfying the condition \(\operatorname{Arg}\left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) is
1 \(\left\{z \in C /\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
2 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6=0\}\)
3 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6>0\), \(\left.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
4 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6\lt 0\), \(\left.\left|z-\frac{1+3 \mathrm{i}}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
Explanation:
C We have, \(\arg \left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) \(\arg (z-1)-\arg (z-3 i)=\frac{\pi}{2}\) \(\arg [(x-1)+i y]-\arg [x+(y-3) i]=\frac{\pi}{2}\) \(\tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y-3}{x}=\frac{\pi}{2}\) \(\tan ^{-1}\left[\frac{\mathrm{y}-1}{1+\frac{y}{x} \times \frac{y}{x}}\right]=\frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(y-3)}=\frac{1}{0}\) \(x(x-1)+y(y-3)=0\) \(x^2+y^2-x-3 y=0\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{1}{4}+\frac{9}{4}\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{\sqrt{10}}{2}\right)^2\) Which is circle with centre \(\left(\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{\sqrt{10}}{2}\) \(\therefore z \in C:(3-i) z+(3+i) \bar{z}-6>0,\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Complex Numbers and Quadratic Equation
117976
In the argand diagram, all the complex numbers \(z\) satisfying \(|z-4 i|+|z+4 i|=10\) lie on a/an
1 straight line
2 circle
3 ellipse
4 parabola
Explanation:
C Given, \(|z-4 i|+|z+4 i|=10\) \(\text { Put, } z=x+i y\) \(\therefore |x+i y-4 i|+|x+i y+4 i|=10\) \(\text { or } |x+i(y-4)|+|x+i(y+4)|=10\) \(\sqrt{x^2+(y-4)^2}+\sqrt{x^2+(y+4)^2}=10\) \(\text { or } \sqrt{x^2+(y-4)^2}=10-\sqrt{x^2+(y+4)^2}\) \(\text { or } x^2+(y-4)^2=100+x^2+(y+4)^2\) \(\quad-20 \cdot \sqrt{x^2+(y+4)^2}\) \(\text { or } 100+x^2+y^2+16+8 y-x^2-y^2-16+8 y\) \(=20 \cdot \sqrt{x^2+(y+4)^2}\) \((100+16 y)=20 \cdot \sqrt{x^2+(y+4)^2}\) \((25+4 y)=5 \sqrt{x^2+(y+4)^2}\) Squaring again both the sides, we get \((25+4 y)^2=25\left(x^2+(y+4)^2\right)\) \((25+4 y)^2=25\left(x^2+\left(y^2+16+8 y\right)\right)\) \((25)^2+16 y^2+200 y=25 x^2+25 y^2+25 \times 16\) \(+25 \times 8 \mathrm{y}\) \(\therefore \quad 625-25 \times 16=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25(25-16)=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(25 \times 9=25 \mathrm{x}^2+9 \mathrm{y}^2\) \(\frac{\mathrm{x}^2}{9}+\frac{\mathrm{y}^2}{25}=1\) which is an ellips.
EAMCET-1996
Complex Numbers and Quadratic Equation
117977
\(P\) represents \(z=x+\) iy in argand plane and \(|3 z-1|=3|z-2|\), then locus of \(P\) is
1 \(x=0\)
2 \(x^2+y^2=8\)
3 \(y=x\)
4 \(6 x=7\)
Explanation:
D \(\mathrm{P}\) represents \(\mathrm{z}=\mathrm{x}+\) iy in the argand plane. And, \(z=x+i y\), Now, \(|3 z-1|=3|z-2|\) \(|3 \mathrm{x}+3 \mathrm{iy}-1|=3|\mathrm{x}+\mathrm{iy}-2|\) or \(\quad|(3 \mathrm{x}-1)+3 \mathrm{y} \cdot \mathrm{i}|=3|(\mathrm{x}-2)+\mathrm{iy}|\) or \(\quad(3 \mathrm{x}-1)^2+(3 \mathrm{y})^2=9\left[(\mathrm{x}-2)^2+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9\left[\mathrm{x}^2+4-4 \mathrm{x}+\mathrm{y}^2\right]\) \(9 \mathrm{x}^2+1-6 \mathrm{x}+9 \mathrm{y}^2=9 \mathrm{x}^2+9 \mathrm{y}^2+36-36 \mathrm{x}\) \(30 \mathrm{x}=35\) or \(\mathrm{x}=\frac{35}{30}=\frac{7}{6}\) \(6 \mathrm{x}=7\)
EAMCET-1999
Complex Numbers and Quadratic Equation
117978
The locus of the \(\mathrm{z}\) in the argand plane for which \(|z+1|^2+|z-1|^2=4\), is a
1 straight line
2 pair of straight lines
3 circle
4 parabola
Explanation:
C Given, \(|\mathrm{z}+1|^2+|\mathrm{z}-1|^2=4\) \(\text { Put, } \mathrm{z}=\mathrm{x}+\mathrm{iy}\) \(\therefore |(\mathrm{x}+1)+\mathrm{iy}|^2+|(\mathrm{x}-1)+\mathrm{iy}|^2=4\) \(\text { or } (\mathrm{x}+1)^2+\mathrm{y}^2+(\mathrm{x}-1)^2+\mathrm{y}^2=4\) \(\text { or } \mathrm{x}^2+1+2 \mathrm{x}+\mathrm{y}^2+\mathrm{x}^2+1-2 \mathrm{x}+\mathrm{y}^2=4\) \(\text { or } 2 \mathrm{x}^2+2 \mathrm{y}^2+2=4\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2+1=2\) \(\text { or } \mathrm{x}^2+\mathrm{y}^2=1 \text { which is a circle. }\)
EAMCET-2000
Complex Numbers and Quadratic Equation
117979
If \(z=x+i y, x, y \in[R\) and if the point \(P\) in the Argand plane represents \(z\), then the locus of \(P\) satisfying the condition \(\operatorname{Arg}\left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) is
1 \(\left\{z \in C /\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
2 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6=0\}\)
3 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6>0\), \(\left.\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
4 \(\{z \in C /(3-i) z+(3+i) \bar{z}-6\lt 0\), \(\left.\left|z-\frac{1+3 \mathrm{i}}{2}\right|=\frac{\sqrt{10}}{2}\right\}\)
Explanation:
C We have, \(\arg \left(\frac{z-1}{z-3 i}\right)=\frac{\pi}{2}\) \(\arg (z-1)-\arg (z-3 i)=\frac{\pi}{2}\) \(\arg [(x-1)+i y]-\arg [x+(y-3) i]=\frac{\pi}{2}\) \(\tan ^{-1} \frac{y}{x-1}-\tan ^{-1} \frac{y-3}{x}=\frac{\pi}{2}\) \(\tan ^{-1}\left[\frac{\mathrm{y}-1}{1+\frac{y}{x} \times \frac{y}{x}}\right]=\frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(x-3)}=\tan \frac{\pi}{2}\) \(\frac{x y-(x-1)(x-3)}{x(x-1)+y(y-3)}=\frac{1}{0}\) \(x(x-1)+y(y-3)=0\) \(x^2+y^2-x-3 y=0\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\frac{1}{4}+\frac{9}{4}\) \(\left(x-\frac{1}{2}\right)^2+\left(y-\frac{3}{2}\right)^2=\left(\frac{\sqrt{10}}{2}\right)^2\) Which is circle with centre \(\left(\frac{1}{2}, \frac{3}{2}\right)\) and radius \(\frac{\sqrt{10}}{2}\) \(\therefore z \in C:(3-i) z+(3+i) \bar{z}-6>0,\left|z-\frac{1+3 i}{2}\right|=\frac{\sqrt{10}}{2}\)