Geometry Practice Test 6

OP = 5 cm
O’P = 5 cm
OA = O’B = r = 3 cm
In ΔOAP,
${\left(\mathit{O}\mathit{P}\right)}^2$ = ${\left(\mathit{O}\mathit{A}\right)}^2+{\left(\mathit{A}\mathit{P}\right)}^2$
${\left(5\right)}^2$ = ${\left(3\right)}^2+{\left(\mathit{A}\mathit{P}\right)}^2$
$\sqrt{25-9}$ = AP

AP = 4 cm
Similarly, we can find PB = 4 cm
So, the length of the transverse common tangent = 8 cm