If $(a, b)$ and $(c, d)$ are the internal and external centres of similitudes of the circles $x^2+y^2+4…
If $(a, b)$ and $(c, d)$ are the internal and external centres of similitudes of the circles $x^2+y^2+4 x-5=0$ and $x^2+y^2$ $-6 y+8=0$ respectively, then $(a+d)(b+c)=$
$4$
$9$
$13$
$22$
Solution
Given equation of two circle
$x^2+y^2+4 x-5=0 \text { and } x^2+y^2-6 y+8=0$
$\Rightarrow(x+2)^2+y^2=3^2 \text { and, } x^2+(y-3)^2=1$
So, $\mathrm{C}_1=(-2,0), r_1=3 ; \mathrm{C}_2=(0,3), r_2=1$
$\begin{aligned} & \text { Now, internal centre }=\left(\frac{r_2 x_1+r_1 x_2}{r_1+r_2}, \frac{r_2 y_1+r_1 y_2}{r_1+r_2}\right) \\ & =\left(\frac{-2+0}{4}, \frac{0+9}{4}\right) \Rightarrow(a, b)=\left(\frac{-1}{2}, \frac{9}{4}\right)\end{aligned}$
$\begin{aligned} & \text { and external centre }=\left(\frac{x_2 r_1-r_2 x_1}{r_1-r_2}, \frac{r_1 y_2-r_2 y_1}{r_1-r_2}\right) \\ & =\left(\frac{0+2}{3-1}, \frac{3 \times 3-0}{3-1}\right) \Rightarrow(\mathrm{c}, \mathrm{d})=\left(1, \frac{9}{2}\right)\end{aligned}$
Now, $(a+d)(b+c)=\left(\frac{-1}{2}+\frac{9}{2}\right)\left(\frac{9}{4}+1\right)=4 \times \frac{13}{4}=13$