From the center O of the plane of the circle, the perpendicular OM = 12 cm is raised. The tangent AC is drawn to the plane

From the center O of the plane of the circle, the perpendicular OM = 12 cm is raised. The tangent AC is drawn to the plane of the circle. Distance MC = 15 cm, OA = r = 6 cm. Find AC.

Consider a triangle OCM – it is rectangular with a right angle at the vertex O.
Because OM is perpendicular to the plane, which means it is perpendicular to any straight line in this plane.
Find the OS leg:
OS = √ (MC ^ 2-MO ^ 2) = √ (225-144) = 9 cm.
Now consider the right-angled triangle OAC. Right angle – A. Because AC is tangent to the circle, and it is always perpendicular to the radius drawn to this tangent.
We know OC and OA, and we can find CA by the Pythagorean theorem.
CA = √ (OC ^ 2-OA ^ 2) = √ (81-36) = 3 (5) cm
Answer 3 (5) see