Segment AB=6 is tangent to the circle with center O. Lengths of other segments are BC=8, CD=10, and OD=12. Find the radius of the circle.
+128475 Let E be the intersection of BC with the circle
By the secant tangent theorem,
BE ( BE + EC) = AB^2
BE ( 8) = 6^2
BE (8) = 36
BE = 36/8 = 18/4 = 4.5
So chord CE = 8 - 4.5 = 3.5
If we call the midpoint of this chord, F and we draw OF....then triangle ODF is right with OD the hypotenuse
And DF = 10 + CF = 10 + CE/2 = 10 + 3.5/2 = 11.75
And by the Pythagorean Theorem, OF^2 = OD^2 - DF^2 = 5.9375
And we can draw another right triangle OFC with the radius of the circle, r, as a hypotenuse, legs CF and OF
So
r^2 = CF^2 + OF^2
r^2 = (1.75)^2 + 5.9375
r^2 = 9
r = 3
