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(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,normalize the equations so that the coef

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$-\frac{1}{10}$

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(B) The equation of the radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$.First,normalize the equations so that the coefficients of ${x^2}$ and ${y^2}$ are $1$.For the first circle: ${S_1} \equiv {x^2} + {y^2} - 3x - 4y + 5 = 0$.For the second circle,divide by $3$: ${S_2} \equiv {x^2} + {y^2} - \frac{7}{3}x + \frac{8}{3}y + \frac{11}{3} = 0$.Subtracting $S_2$ from $S_1$:$({x^2} + {y^2} - 3x - 4y + 5) - ({x^2} + {y^2} - \frac{7}{3}x + \frac{8}{3}y + \frac{11}{3}) = 0$$(-3 + \frac{7}{3})x + (-4 - \frac{8}{3})y + (5 - \frac{11}{3}) = 0$$-\frac{2}{3}x - \frac{20}{3}y + \frac{4}{3} = 0$Multiplying by $-3$: $2x + 20y - 4 = 0$,which simplifies to $x + 10y - 2 = 0$.The equation of the line is $10y = -x + 2$,or $y = -\frac{1}{10}x + \frac{1}{5}$.The gradient (slope) is $-\frac{1}{10}$.

The gradient of the radical axis of the circles ${x^2} + {y^2} - 3x - 4y + 5 = 0$ and $3{x^2} + 3{y^2} - 7x + 8y + 11 = 0$ is

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