Let complex numbers and 1 lie on circles ( x - x 0 ) 2 + ( y - y 0 ) 2 = r 2 and ( x - x 0 ) 2 + ( y - y 0 ) 2 = 4 r 2 respectively. If z 0 = x 0 + i y 0 satisfies the equation 2 | z 0 | 2 = r 2 + 2 , then | | = [2013]

Question

Let complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on circles ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$ and ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$ respectively. If $z_{0} = x_{0} + i y_{0}$ satisfies the equation $2$ $| z_{0} |^{2}$ $= r^{2} + 2$ , then $| α |$ $=$ [2013]

Smart Achievers · Accepted Answer

(3)

$Since, α lies on the circle {(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$

$∴$ $| α - z_{0} |^{2}$ $= r^{2}$

$\Rightarrow (α - z_{0}) (\bar{α} - \bar{z_{0}}) = r^{2}$

$\Rightarrow α \bar{α} - α \bar{z_{0}} - \bar{α} z_{0} + z_{0} \bar{z_{0}} = r^{2}$

$\Rightarrow | α |^{2} + | z_{0} |^{2} - α \bar{z_{0}} - \bar{α} z_{0} = r^{2} . . . (i)$

$Also \frac{1}{\bar{α}} lies on the circle {(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$

$∴$ $| \frac{1}{\bar{α}} - z_{0} |^{2}$ $= 4 r^{2}$ $\Rightarrow (\frac{1}{\bar{α}} - z_{0}) (\frac{1}{α} - \bar{z_{0}}) = 4 r^{2}$

$\Rightarrow \frac{1}{α \bar{α}} - \frac{z_{0}}{α} - \frac{\bar{z_{0}}}{\bar{α}} + z_{0} \bar{z_{0}} = 4 r^{2}$

$\Rightarrow \frac{1}{| α |^{2}} - \frac{z_{0} \bar{α}}{| α |^{2}} - \frac{\bar{z_{0}} α}{| α |^{2}} + | z_{0} |^{2} = 4 r^{2}$

$\Rightarrow 1 + | α |^{2} | z_{0} |^{2} - z_{0} \bar{α} - \bar{z_{0}} α = 4 r^{2} | α |^{2} . . . (i i)$

$On subtracting equation (i) from (ii), we get$

$1 - | α |^{2} + | z_{0} |^{2} (| α |^{2} - 1) = r^{2} (4 | α |^{2} - 1)$

$or (| α |^{2} - 1) (| z_{0} |^{2} - 1) = r^{2} (4 | α |^{2} - 1)$

$Using | z_{0} |^{2} = \frac{r^{2} + 2}{2}, we get$

$(| α |^{2} - 1) \frac{r^{2}}{2} = r^{2} (4 | α |^{2} - 1)$

$\Rightarrow | α |^{2} - 1 = 8 | α |^{2} - 2$

$\Rightarrow | α |^{2} = \frac{1}{7}$

$\Rightarrow | α | = \frac{1}{\sqrt{7}}$

Solution

Q.
Let complex numbers $α$ and $\frac{1}{\bar{α}}$ lie on circles ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = r^{2}$ and ${(x - x_{0})}^{2} + {(y - y_{0})}^{2} = 4 r^{2}$ respectively. If $z_{0} = x_{0} + i y_{0}$ satisfies the equation $2$ $| z_{0} |^{2}$ $= r^{2} + 2$ , then $| α |$ $=$ [2013]

1 $\frac{1}{\sqrt{2}}$

2 $\frac{1}{2}$

3 $\frac{1}{\sqrt{7}}$

4 $\frac{1}{3}$

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Solution

Q. Let complex numbers α and 1α¯ lie on circles (x-x0)2+(y-y0)2=r2 and (x-x0)2+(y-y0)2=4r2 respectively. If z0=x0+iy0 satisfies the equation 2|z0|2=r2+2, then |α|= [2013]

1 12

2 12

3 17