Geometry Practice Test 4

∠BGC = 2x = 90°
Let r1 is the radius of the bigger circle = 5 cm.
${\mathit{r}}^2$ is radius of the smaller circle = 3 cm
Distance between both the centers of the circle
d = 24
x = $\frac{\mathit{d}{\mathit{r}}_2}{{\mathit{r}}_1+{\mathit{r}}_2}$
x = $\frac{\left(24\right).\left(3\right)}{5+5}$ = $\frac{72}{8}$ = 9

x = 9
By using the length of the transversal tangent
= $\sqrt{{\left(\mathit{d}-\mathit{x}\right)}^2-\mathit{r}{1}^2}$ + $\sqrt{{\mathit{x}}^2-{\mathit{r}}_2^2}$ = l
l = $\sqrt{{\left(2-9\right)}^2-{\left(5\right)}^2}$ + $\sqrt{{\left(9\right)}^2-{\left(3\right)}^2}$
= $\sqrt{255-25+\sqrt{81-9}}$
= $\sqrt{200}+\sqrt{72}$
= $16\sqrt{2}$