Let the complex numbers be represented in the complex plane where the center of the circles is z 0 = x 0 + i y 0 z_0 = x_0 + iy_0 z 0 = x 0 + i y 0
The equations of the two circles in terms of complex numbers are:
For α \alpha α ( x − x 0 ) 2 + ( y − y 0 ) 2 = r 2 (x - x_0)^2 + (y - y_0)^2 = r^2 ( x − x 0 ) 2 + ( y − y 0 ) 2 = r 2 ∣ α − z 0 ∣ 2 = r 2 |\alpha - z_0|^2 = r^2 ∣ α − z 0 ∣ 2 = r 2
For 1 α ˉ \frac{1}{\bar{\alpha}} α ˉ 1 ( x − x 0 ) 2 + ( y − y 0 ) 2 = 4 r 2 (x - x_0)^2 + (y - y_0)^2 = 4r^2 ( x − x 0 ) 2 + ( y − y 0 ) 2 = 4 r 2 ∣ 1 α ˉ − z 0 ∣ 2 = 4 r 2 \left|\frac{1}{\bar{\alpha}} - z_0\right|^2 = 4r^2 α ˉ 1 − z 0 2 = 4 r 2
Expanding equation (1):
( α − z 0 ) ( α ˉ − z ˉ 0 ) = r 2 (\alpha - z_0)(\bar{\alpha} - \bar{z}_0) = r^2 ( α − z 0 ) ( α ˉ − z ˉ 0 ) = r 2 ∣ α ∣ 2 − ( α z ˉ 0 + α ˉ z 0 ) + ∣ z 0 ∣ 2 = r 2 |\alpha|^2 - (\alpha \bar{z}_0 + \bar{\alpha} z_0) + |z_0|^2 = r^2 ∣ α ∣ 2 − ( α z ˉ 0 + α ˉ z 0 ) + ∣ z 0 ∣ 2 = r 2 α z ˉ 0 + α ˉ z 0 = ∣ α ∣ 2 + ∣ z 0 ∣ 2 − r 2 — (3) \alpha \bar{z}_0 + \bar{\alpha} z_0 = |\alpha|^2 + |z_0|^2 - r^2 \quad \text{--- (3)} α z ˉ 0 + α ˉ z 0 = ∣ α ∣ 2 + ∣ z 0 ∣ 2 − r 2 — (3)
Expanding equation (2):
∣ 1 − z 0 α ˉ α ˉ ∣ 2 = 4 r 2 \left|\frac{1 - z_0 \bar{\alpha}}{\bar{\alpha}}\right|^2 = 4r^2 α ˉ 1 − z 0 α ˉ 2 = 4 r 2 ∣ 1 − z 0 α ˉ ∣ 2 ∣ α ∣ 2 = 4 r 2 \frac{|1 - z_0 \bar{\alpha}|^2}{|\alpha|^2} = 4r^2 ∣ α ∣ 2 ∣1 − z 0 α ˉ ∣ 2 = 4 r 2 ∣ 1 − z 0 α ˉ ∣ 2 = 4 r 2 ∣ α ∣ 2 |1 - z_0 \bar{\alpha}|^2 = 4r^2 |\alpha|^2 ∣1 − z 0 α ˉ ∣ 2 = 4 r 2 ∣ α ∣ 2 ( 1 − z 0 α ˉ ) ( 1 − z ˉ 0 α ) = 4 r 2 ∣ α ∣ 2 (1 - z_0 \bar{\alpha})(1 - \bar{z}_0 \alpha) = 4r^2 |\alpha|^2 ( 1 − z 0 α ˉ ) ( 1 − z ˉ 0 α ) = 4 r 2 ∣ α ∣ 2 1 − ( α z ˉ 0 + α ˉ z 0 ) + ∣ z 0 ∣ 2 ∣ α ∣ 2 = 4 r 2 ∣ α ∣ 2 — (4) 1 - (\alpha \bar{z}_0 + \bar{\alpha} z_0) + |z_0|^2 |\alpha|^2 = 4r^2 |\alpha|^2 \quad \text{--- (4)} 1 − ( α z ˉ 0 + α ˉ z 0 ) + ∣ z 0 ∣ 2 ∣ α ∣ 2 = 4 r 2 ∣ α ∣ 2 — (4)
Substituting equation (3) into equation (4):
1 − ( ∣ α ∣ 2 + ∣ z 0 ∣ 2 − r 2 ) + ∣ z 0 ∣ 2 ∣ α ∣ 2 = 4 r 2 ∣ α ∣ 2 1 - \left(|\alpha|^2 + |z_0|^2 - r^2\right) + |z_0|^2 |\alpha|^2 = 4r^2 |\alpha|^2 1 − ( ∣ α ∣ 2 + ∣ z 0 ∣ 2 − r 2 ) + ∣ z 0 ∣ 2 ∣ α ∣ 2 = 4 r 2 ∣ α ∣ 2 1 − ∣ α ∣ 2 − ∣ z 0 ∣ 2 + ∣ z 0 ∣ 2 ∣ α ∣ 2 + r 2 = 4 r 2 ∣ α ∣ 2 1 - |\alpha|^2 - |z_0|^2 + |z_0|^2 |\alpha|^2 + r^2 = 4r^2 |\alpha|^2 1 − ∣ α ∣ 2 − ∣ z 0 ∣ 2 + ∣ z 0 ∣ 2 ∣ α ∣ 2 + r 2 = 4 r 2 ∣ α ∣ 2 ( 1 − ∣ α ∣ 2 ) ( 1 − ∣ z 0 ∣ 2 ) + r 2 = 4 r 2 ∣ α ∣ 2 (1 - |\alpha|^2)(1 - |z_0|^2) + r^2 = 4r^2 |\alpha|^2 ( 1 − ∣ α ∣ 2 ) ( 1 − ∣ z 0 ∣ 2 ) + r 2 = 4 r 2 ∣ α ∣ 2
We are given the condition:
2 ∣ z 0 ∣ 2 = r 2 + 2 ⟹ 1 − ∣ z 0 ∣ 2 = − r 2 2 2|z_0|^2 = r^2 + 2 \implies 1 - |z_0|^2 = -\frac{r^2}{2} 2∣ z 0 ∣ 2 = r 2 + 2 ⟹ 1 − ∣ z 0 ∣ 2 = − 2 r 2
Substituting 1 − ∣ z 0 ∣ 2 = − r 2 2 1 - |z_0|^2 = -\frac{r^2}{2} 1 − ∣ z 0 ∣ 2 = − 2 r 2 ( 1 − ∣ α ∣ 2 ) ( − r 2 2 ) + r 2 = 4 r 2 ∣ α ∣ 2 (1 - |\alpha|^2)\left(-\frac{r^2}{2}\right) + r^2 = 4r^2 |\alpha|^2 ( 1 − ∣ α ∣ 2 ) ( − 2 r 2 ) + r 2 = 4 r 2 ∣ α ∣ 2
Assuming r ≠ 0 r \neq 0 r = 0 r 2 r^2 r 2 − 1 2 ( 1 − ∣ α ∣ 2 ) + 1 = 4 ∣ α ∣ 2 -\frac{1}{2}(1 - |\alpha|^2) + 1 = 4|\alpha|^2 − 2 1 ( 1 − ∣ α ∣ 2 ) + 1 = 4∣ α ∣ 2 − 1 2 + 1 2 ∣ α ∣ 2 + 1 = 4 ∣ α ∣ 2 -\frac{1}{2} + \frac{1}{2}|\alpha|^2 + 1 = 4|\alpha|^2 − 2 1 + 2 1 ∣ α ∣ 2 + 1 = 4∣ α ∣ 2 1 2 = ( 4 − 1 2 ) ∣ α ∣ 2 \frac{1}{2} = \left(4 - \frac{1}{2}\right)|\alpha|^2 2 1 = ( 4 − 2 1 ) ∣ α ∣ 2 1 2 = 7 2 ∣ α ∣ 2 \frac{1}{2} = \frac{7}{2}|\alpha|^2 2 1 = 2 7 ∣ α ∣ 2 ∣ α ∣ 2 = 1 7 |\alpha|^2 = \frac{1}{7} ∣ α ∣ 2 = 7 1 ∣ α ∣ = 1 7 |\alpha| = \frac{1}{\sqrt{7}} ∣ α ∣ = 7 1
Therefore, the correct option is (C) .