Let a circle C in the complex plane pass through the points z_1=3+4i, z_2=4+3i, z_3=5i.

Question

Let a circle $C$ in the complex plane pass through the points $z_1=3+4i,\ z_2=4+3i,\ z_3=5i$. If $z(
eq z_1)$ is a point on $C$ such that the line through $z$ and $z_1$ is perpendicular to the line through $z_2$ and $z_3$, then $\arg(z)$ is equal to:

Accepted Answer

Finding the circle through $(3,4),(4,3),(0,5)$: substituting into $x^2+y^2+Dx+Ey+F=0$ and solving gives $D=E=0,\ F=-25$, i.e. $x^2+y^2=25$ --- the circle $|z|=5$.

The chord direction $z_2-z_3=(4+3i)-5i=4-2i$. We need $(z-z_1)$ to be a purely imaginary multiple of $4-2i$, i.e.
$$\operatorname{Re}\left(\frac{(z-z_1)}{4-2i}\right)=0.$$

With $z=x+iy,\ z_1=3+4i$: this simplifies to $4(x-3)-2(y-4)=0 \implies y=2x-2$.

Combined with $x^2+y^2=25$: $x^2+(2x-2)^2=25 \implies 5x^2-8x-21=0 \implies x=3$ or $x=-\dfrac75$.

$x=3$ gives $z=z_1$ (excluded). $x=-\dfrac75$ gives $y=-\dfrac{24}5$, so $z=-\dfrac75-\dfrac{24}5i$.

This point lies in the third quadrant, so:
$$\arg(z)=-\pi+\tan^{-1}\left(\frac{24/5}{7/5}\right)=\tan^{-1}\frac{24}7-\pi.$$

Solutions · 1