From a point outside the circle, a tangent is drawn, with a length of 12, and the largest secant with a length of 24

From a point outside the circle, a tangent is drawn, with a length of 12, and the largest secant with a length of 24. Find the shortest distance from this point to the points of the circumference of this circle.

First of all, it should be noted that the largest secant will be the secant that passes through the center of the circle. In addition, it is obvious that the part of this secant from a given point to a circle will be the shortest distance from a point to points on the circumference of a circle. So, A is a point outside the circle; AB – tangent; АС – the largest secant; AD – the shortest distance from point A to the points of the circumference of this circle. You need to define AD.
Let’s introduce the notation: AD = x and radius OB = OC = OD = r. We have AC = OC + OD + AD = 2 * r + x = 24.
Since the tangent is perpendicular to the radius of the circle drawn to the point of tangency, then OB ⊥ AB. Hence, ABO is a right-angled triangle, where AB = 12 and OB = r are legs, and OA = OD + AD = r + x is the hypotenuse. By the Pythagorean theorem, OA ^ 2 = AB ^ 2 + OB ^ 2, that is, (r + x) ^ 2 = 12 ^ 2 + r ^ 2 or r ^ 2 + 2 * x * r + x ^ 2 = 144 + r ^ 2, whence 2 * x * r + x2 = 144. Taking the factor x out of the brackets, we have (2 * r + x) * x = 144 or (taking into account the last equality of item 2) 24 * x = 144, whence x = 144: 24 = 6.
Answer: 6.