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(C) Given equations of the two circles are:$x^2+y^2+4x-5=0 \Rightarrow (x+2)^2+y^2=3^2$$x^2+y^2-6y+8=0 \Rightarrow x^2+(y-3)^2=1$Thus,$C_1=(-2, 0), r_

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$13$

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(C) Given equations of the two circles are:$x^2+y^2+4x-5=0 \Rightarrow (x+2)^2+y^2=3^2$$x^2+y^2-6y+8=0 \Rightarrow x^2+(y-3)^2=1$Thus,$C_1=(-2, 0), r_1=3$ and $C_2=(0, 3), r_2=1$.The internal centre of similitude is given by $\left(\frac{r_2x_1+r_1x_2}{r_1+r_2}, \frac{r_2y_1+r_1y_2}{r_1+r_2}ight)$:$(a, b) = \left(\frac{1(-2)+3(0)}{3+1}, \frac{1(0)+3(3)}{3+1}ight) = \left(-\frac{2}{4}, \frac{9}{4}ight) = \left(-\frac{1}{2}, \frac{9}{4}ight)$.The external centre of similitude is given by $\left(\frac{r_1x_2-r_2x_1}{r_1-r_2}, \frac{r_1y_2-r_2y_1}{r_1-r_2}ight)$:$(c, d) = \left(\frac{3(0)-1(-2)}{3-1}, \frac{3(3)-1(0)}{3-1}ight) = \left(\frac{2}{2}, \frac{9}{2}ight) = \left(1, \frac{9}{2}ight)$.Now,calculating $(a+d)(b+c)$:$(a+d)(b+c) = \left(-\frac{1}{2} + \frac{9}{2}ight) \left(\frac{9}{4} + 1ight) = \left(\frac{8}{2}ight) \left(\frac{13}{4}ight) = 4 	imes \frac{13}{4} = 13$.

If $(a, b)$ and $(c, d)$ are the internal and external centres of similitude of the circles $x^2+y^2+4x-5=0$ and $x^2+y^2-6y+8=0$ respectively,then $(a+d)(b+c)=$

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