The distance of the origin from the external centre of similitude for the circles $x^2+y^2-8 x-10 y-8=0$ and…
The distance of the origin from the external centre of similitude for the circles $x^2+y^2-8 x-10 y-8=0$ and $\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{x}-2 \mathrm{y}-2=0$ is
$\frac{3 \sqrt{26}}{5}$
$\frac{\sqrt{290}}{9}$
$\frac{\sqrt{290}}{5}$
$\frac{\sqrt{26}}{3}$
Solution
Let $\mathrm{S}_1: x^2+y^2-8 x-10 y-8=0$
$\begin{aligned} & \mathrm{C}_1=(4,5) \& r_1=\sqrt{16+25+8}=7 \\ & \text { and } \mathrm{S}_2: x^2+y^2+2 x-2 y-2=0 \\ & \mathrm{C}_2=(-1,1) \& r_2=\sqrt{1+1+2}=2\end{aligned}$
Using section formula, external centre of similitude is:
$\mathrm{Q} \equiv\left(\frac{r_1 x_2-x_1 r_2}{r_1-r_2}, \frac{r_1 y_2-r_2 y_1}{r_1-r_2}\right)$
$\equiv\left(\frac{-7 \times 1-4 \times 2}{7-2}, \frac{7 \times 1-2 \times 5}{7-2}\right) \equiv\left(-3,-\frac{3}{5}\right)$
Distance from origin to $\mathrm{Q}$ is
$\mathrm{D}=\sqrt{(-3)^2+\left(\frac{-3}{5}\right)^2}=\sqrt{1+\frac{1}{25}} \cdot(3)=\frac{3}{5} \sqrt{26}$