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Given $z_1, z_2 \in \mathbb{C}$, where $z_1 \neq z_2$, and $\lambda \in (\mathbb{R}^+ \cup \{0\}) \setminus \{1\}$, prove that the equation $|\frac{z-z_1}{z-z_2}| = \lambda$ determines a circle and find out the center and radius of this circle.

My notes say that for a circle with center $z_0$ and radius $r$, it can be formed by the graphical representation of the equation $|z-z_0|^2 = r^2$.

What I attempted was to expand each complex number $z_k$ into $\Re(z_k) + i\Im(z_k) = x_k + iy_k$ until I eventually got a large expression that itself is in the form $z - z_0$ , but I probably made silly algebraic errors on the way, getting the following (likely wrong) results:

The center of the circle determined is $(\frac{x(x_1+x_2) + y(y_1+y_2)}{(x-x_2)^2 + (y-y_2)^2}, \frac{xy_1+yx_2+x_1y_2}{(x-x_2)^2+(y-y_2)^2})$ and it has a radius of length $\lambda$. However, I am not sure how right this is, because $x$ and $y$ themselves are allowed to vary, meaning that the center of this circle, while it has a radius constant at $\lambda$, could change depending on how I split the $|z-z_0|$-form expression I obtained originally.

Thank you!

chuloobydooby
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3 Answers3

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Note that $|z|^2=z\bar z$.

\begin{align*} |z-z_1|^2&=\lambda^2|z-z_2|^2\\ (z-z_1)(\bar{z}-\bar{z_1})&=\lambda^2(z-z_2)(\bar{z}-\bar{z_2})\\ z\bar{z}-z_1\bar{z}-\bar{z_1}z+z_1\bar{z_1}&=\lambda^2(z\bar{z}-z_2\bar{z}-\bar{z_2}z+z_2\bar{z_2})\\ (\lambda^2-1)z\bar{z}-(\lambda^2z_2-z_1)\bar{z}-(\lambda^2\bar{z_2}-\bar{z_1})z&=z_1\bar{z_1}-\lambda^2z_2\bar{z_2}\\ z\bar{z}-\frac{\lambda^2z_2-z_1}{\lambda^2-1}\bar{z}-\frac{\lambda^2\bar{z_2}-\bar{z_1}}{\lambda^2-1}z&=\frac{z_1\bar{z_1}-\lambda^2z_2\bar{z_2}}{\lambda^2-1}\\ \left(z-\frac{\lambda^2z_2-z_1}{\lambda^2-1}\right)\left(\bar{z}-\frac{\lambda^2\bar{z_2}-\bar{z_1}}{\lambda^2-1}\right)&=\frac{z_1\bar{z_1}-\lambda^2z_2\bar{z_2}}{\lambda^2-1}+\frac{(\lambda^2z_2-z_1)(\lambda^2\bar{z_2}-\bar{z_1})}{(\lambda^2-1)^2}\\ \left(z-\frac{\lambda^2z_2-z_1}{\lambda^2-1}\right)\left(\bar{z}-\frac{\lambda^2\bar{z_2}-\bar{z_1}}{\lambda^2-1}\right)&=\frac{(\lambda^2-1)z_1\bar{z_1}-\lambda^2(\lambda^2-1)z_2\bar{z_2}+\lambda^4z_2\bar{z_2}+z_1\bar{z_1}-\lambda^2z_1\bar{z_2}-\lambda^2\bar{z_1}z_2}{(\lambda^2-1)^2}\\ \left|z-\frac{\lambda^2z_2-z_1}{\lambda^2-1}\right|^2&=\frac{\lambda^2z_1\bar{z_1}+\lambda^2z_2\bar{z_2}-\lambda^2z_1\bar{z_2}-\lambda^2\bar{z_1}z_2}{(\lambda^2-1)^2} \end{align*}

It represents a circle with centre at $\dfrac{\lambda^2z_2-z_1}{\lambda^2-1}$ and radius $\dfrac{\sqrt{\lambda^2(|z_1|^2+|z_2|^2)-2\lambda^2\Re(z_1\bar{z_2})}}{|\lambda^2-1|}$.

CY Aries
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Perhaps it is easiest to think about the equation in the Cartesian plane. Taking $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, and then squaring both sides and collecting like terms, we are left with: $$(1-\lambda^2)x^2+(1-\lambda^2)y^2+2(\lambda^2 x_2-x_1)x+2(\lambda^2 y_2 -y_1) y+(x_1^2+y_1^2-\lambda(x_2^2+y_2^2))=0$$ This shows very clearly that the equation represents a circle, because, comparing with a general conic: $$ax^2+2hxy+by^2+2gx+2fy+c=0$$ We see that here, $a=b=(1-\lambda ^2)$ and $h=0$, moreover it can be verified that $\Delta \neq 0$. Notice that this also shows that $\lambda\neq 1$ is necessary for obtaining a circle, since, if $\lambda=1$ the equation reduces to: $$2(x_2-x_1)x+2(y_2-y_1)y+(x_1^2+y_1^2-(x_2^2+y_2^2))=0$$ Which represents the perpendicular bisector of the line joining the two points, as expected. Otherwise, it is simple enough to identify centre as: $$(\frac {x_1-\lambda^2 x_2}{1-\lambda^2}, \frac {y_1-\lambda^2 y_2}{1-\lambda^2})$$ which is equivalent to $C: \frac {z_1-\lambda^2 z_2}{1-\lambda^2}$ in the complex plane. The radius is identified as: $$r=\frac{\sqrt {(x_2^2+y_2^2)(\lambda^4-\lambda^3+\lambda)-2\lambda^2(x_1x_2+y_1y_2)}}{1-\lambda^2}$$ A simple bit of algebraic manipulation reduces this to: $$r=\frac {\sqrt {(\lambda^4-\lambda^3+\lambda)|z_2|^2 -\lambda^2(\overline{z_1}z_2+\overline {z_2}z_1)}}{1-\lambda^2}$$ Hence, we are done.

Ritam_Dasgupta
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All you need to do is multiply it out on each side, then move to one side of the equation, and do standard completing the square algebra. Let $z=x+iy$, $z_1=x_1+iy_1$, $z_2=x_2+iy_2$

Multiplying through your equation by the denominator gets us to $$|z-z_1|=\lambda|z-z_2|$$ Squaring both sides and substituting in the values for $x$ and $y$ gets us to $$(x-x_1)^2 + (y-y_1)^2=\lambda^2(x-x_2)^2 + \lambda^2(y-y_2)^2 $$ Multiplying out we get $$x^2 -2x_1x +x_1^2 +y^2 -2y_1y +y_1^2=\lambda^2 x^2 -2\lambda^2x_2x+ \lambda^2x_2^2+\lambda^2 y^2 -2\lambda^2y_2y+ \lambda^2y_2^2$$

Moving the terms with $x^2, y^2, x,y$ to the left and factoring them out and moving the constants to the right we get

$$(1-\lambda^2)x^2 -2(x_1 -\lambda_2x_2)x+(1-\lambda^2)y^2 -2(y_1 -\lambda_2y_2)y=\lambda_2x_2^2+\lambda_2y_2^2-x_1^2 -y_1^2 $$

All that's left is to divide by $1-\lambda^2$, which we know is nonzero by assumption, then complete the square on the left hand side to find the center and the radius.

Alan
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