Point C
lies in the exterior of circle O.
Segment WC
is tangent to circle O
at point W.
CO¯¯¯¯¯¯¯¯,
CW¯¯¯¯¯¯¯¯¯,
and OW¯¯¯¯¯¯¯¯¯
are drawn in to create △OWC.
Circle O as described, where segment O W is labeled r.
If OW=9
and WC=12,
what is CO?
Enter the correct value.
Using the tangent-secant theorem, we know that the square of the length of the tangent segment (WC) is equal to the product of the lengths of the secant segment (OW) and the external segment (CO).
So, we can set up the following equation:
WC^2 = OW * CO
12^2 = 9 * CO
144 = 9 * CO
CO = 144 / 9
CO = 16
Therefore, CO = 16.