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(B) Let $C_1$ and $C_2$ be the centers of the two circles. Since $BP$ and $BQ$ are diameters,$C_1$ is the midpoint of $BP$ and $C_2$ is the midpoint o

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(B) Let $C_1$ and $C_2$ be the centers of the two circles. Since $BP$ and $BQ$ are diameters,$C_1$ is the midpoint of $BP$ and $C_2$ is the midpoint of $BQ$.In $	riangle BPQ$,$C_1$ and $C_2$ are the midpoints of sides $BP$ and $BQ$ respectively. By the midpoint theorem,$PQ \parallel C_1C_2$.The line joining the centers $C_1C_2$ is perpendicular to the radical axis $AB$.Therefore,$PQ$ is perpendicular to the radical axis $AB$.If the slope of the radical axis is $m_1 = 3/4$ and the slope of $PQ$ is $m_2 = a/b$,then $m_1 	imes m_2 = -1$.$\frac{3}{4} 	imes \frac{a}{b} = -1$$\frac{3a}{4b} = -1$$3a = -4b$$3a + 4b = 0$.

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