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(B) The points are $A(3, 4)$,$B(4, 3)$,and $C(0, 5)$.The equation of the circle passing through these points is $x^2 + y^2 = 25$.The slope of the line

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$	an^{-1}\left(\frac{24}{7}ight)-\pi$

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(B) The points are $A(3, 4)$,$B(4, 3)$,and $C(0, 5)$.The equation of the circle passing through these points is $x^2 + y^2 = 25$.The slope of the line segment $BC$ is $m_{BC} = \frac{3-5}{4-0} = \frac{-2}{4} = -\frac{1}{2}$.Since the line through $z(x, y)$ and $z_{1}(3, 4)$ is perpendicular to $BC$,its slope $m$ must satisfy $m 	imes (-1/2) = -1$,so $m = 2$.The equation of this line is $y - 4 = 2(x - 3)$,which simplifies to $y = 2x - 2$.Substituting $y = 2x - 2$ into the circle equation $x^2 + y^2 = 25$:$x^2 + (2x - 2)^2 = 25$$x^2 + 4x^2 - 8x + 4 = 25$$5x^2 - 8x - 21 = 0$$(5x + 7)(x - 3) = 0$.Since $z 
eq z_{1}$,we have $x = -7/5$.Then $y = 2(-7/5) - 2 = -14/5 - 10/5 = -24/5$.Thus,$z = -7/5 - i(24/5)$.Since $z$ is in the third quadrant,$\arg(z) = 	an^{-1}\left(\frac{-24/5}{-7/5}ight) - \pi = 	an^{-1}\left(\frac{24}{7}ight) - \pi$.

Let a circle $C$ in the complex plane pass through the points $z_{1}=3+4i$,$z_{2}=4+3i$,and $z_{3}=5i$. If $z(\neq z_{1})$ 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

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