$ABC$ is a circle with centre O. P is an external point in the line $AB$. From P, a tangent $PC$ has been drawn. If $AB = 10\ cm, BP = 8\ cm$, what is the measure of tangent $PC$?

Since AB is the diameter of 10 cm, the radius is 5 cm and the center O sits on the line AB. Given BP is 8 cm, the distance from the center to the external point is OP = OB + BP = 5 + 8 = 13 cm. Using the tangent-secant theorem, we write PC squared equals OP squared minus the radius squared, which gives PC squared equals 169 minus 25. Solving this yields PC squared equals 144, so the tangent PC is 12 cm.

• Thanks
• Wrong explanation