C_1 is the circle with centre at O(0,0) and r | vedclass

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(B) The equation of circle $C_1$ is $x^2+y^2=16$. The equation of circle $C_2$ is $(x-\alpha)^2+(y-\beta)^2=25$,which expands to $x^2-2\alpha x+\alpha

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$\frac{3}{5}$

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(B) The equation of circle $C_1$ is $x^2+y^2=16$. The equation of circle $C_2$ is $(x-\alpha)^2+(y-\beta)^2=25$,which expands to $x^2-2\alpha x+\alpha^2+y^2-2\beta y+\beta^2=25$. The equation of the common chord is given by $C_2-C_1=0$,which is $-2\alpha x-2\beta y+\alpha^2+\beta^2=9$. The slope of the common chord is $m = -\frac{-2\alpha}{-2\beta} = -\frac{\alpha}{\beta} = \frac{3}{4}$. Let $\alpha = -3\lambda$ and $\beta = 4\lambda$. For the common chord to have maximum length,it must be the diameter of the smaller circle $C_1$,meaning the chord must pass through the centre of $C_1$,which is $(0,0)$. Substituting $(0,0)$ into the chord equation: $-2\alpha(0)-2\beta(0)+\alpha^2+\beta^2=9$,so $\alpha^2+\beta^2=9$. Substituting $\alpha$ and $\beta$ in terms of $\lambda$: $(-3\lambda)^2+(4\lambda)^2=9$ $\Rightarrow 25\lambda^2=9$ $\Rightarrow \lambda = \pm \frac{3}{5}$. If $\lambda = \frac{3}{5}$,then $\alpha = -\frac{9}{5}$ and $\beta = \frac{12}{5}$,so $\alpha+\beta = \frac{3}{5}$. If $\lambda = -\frac{3}{5}$,then $\alpha = \frac{9}{5}$ and $\beta = -\frac{12}{5}$,so $\alpha+\beta = -\frac{3}{5}$.

$C_1$ is the circle with centre at $O(0,0)$ and radius $4$,$C_2$ is a variable circle with centre at $(\alpha, \beta)$ and radius $5$. If the common chord of $C_1$ and $C_2$ has slope $\frac{3}{4}$ and is of maximum length,then one of the possible values of $\alpha+\beta$ is

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