x = 15
Denote the origin of the circle O and the points A, B, C clockwise starting from the left. From the isosceles triangle OAB we get 2 r sin(alpha/2) = 24, where alpha is the angle between OA and OB.
Construct the line orthogonal to OB that goes through C. The length of the line, h, between C and the intersection is h = 7 sin(beta) = x sin(90 - alpha). Denote the lengths of the parts of OB a and b, where a is connected to B. We have a + b = r
Use Thales circle theorem to find that the triangle ABA' completes the red shape, with A' on the circle opposite to A. That means that the angle between A'A and A'B is alpha/2, but A'OB is also an isosceles triangle. So the angle on the other side, beta, has to be the same. Thus, beta = alpha/2.
Now, put everything together: a = 7 cos (alpha/2), b = h cot(90 - alpha) = 7 sin(alpha/2) tan(alpha), r = 12 / sin(alpha/2).
a + b = r <=> cos(alpha/2) sin(alpha/2) + sin^2(alpha/2) tan(alpha) = 12 / 7
1/2 sin(alpha) + 1/2(1 - cos(alpha)) tan(alpha) = 12/7 <=> tan(alpha) = 24/7
From the identity for h we know that x = 7 sin(alpha/2) / cos(alpha). Insert alpha = arctan(24/7)
Yes