The radius of the circle with the center O is 85. The length of the chord AB is 102.

The radius of the circle with the center O is 85. The length of the chord AB is 102. Find the distance from the chord AB to the tangent K parallel to it.

Given: a circle, where ОВ = OK = 85, PQ is the tangent line to the circle centered at point О, AB = 102 and PQ || AB. It is necessary to find the distance from the chord AB to the tangent parallel to it.
As you know, the tangent to the circle is perpendicular to the radius drawn to the tangent point. Therefore OK ⊥ PQ. By the terms of the task, PQ || AB. Therefore, OK ⊥ AB. Hence, ∠ОСВ = 90 °. Then, ΔОСВ is a right-angled triangle.
Obviously, point C is the middle of segment AB, since the ends of segment AB are equidistant from point O and OK ⊥ AB, therefore, segment OK for an isosceles triangle AOB is both the height and the median and bisector. Therefore, CB = AB: 2 = 102: 2 = 51. At the request of the assignment, we find SK. Obviously CK = OK – OC.
By the Pythagorean theorem, OB2 = OS2 + CB2, whence OS2 = OB2 – CB2 = 852 – 512 = 4624. Therefore, OS = 68.
So CK = OK – OC = 85 – 68 = 17.
Answer: the distance from the chord AB to the tangent parallel to it is 17.