An isosceles trapezoid ABCD with bases AD and BC is circumscribed about a circle. Side CD touches this circle

An isosceles trapezoid ABCD with bases AD and BC is circumscribed about a circle. Side CD touches this circle at point Q, and segment AQ intersects the circle at point P. Find the radius of the circle if it is known that AP = 2, PQ = 7

Line segment AQ is the secant of the circle and AN is the tangent line. By the secant and tangent theorem:
AN ^ 2 = AP * AQ = AP * (AP + PQ) = 2 * 9 = 18.
AN = √18.
Let L be the tangent point of the base AD with the circle. But by the property of the trapezoid into which the circle is inscribed, and taking into account that the trapezoid is isosceles, then AN = AL = LD = QD = √18.
Consider a triangle AQD, AD = 2AL = 2√18, AQ = AP + PQ = 9.
Hence, by the cosine theorem:
Cos (AQD) = (AD ^ 2 + AQ ^ 2 – QD ^ 2) / (2 * AD * AQ) = (4 * 18 + 81 – 18) / (2 * 2√18 * 9) = 5 / ( 4√2).
α = arcos (5 / (4√2).
By the property of a tangent, the tangent is perpendicular to the radius.
CD – tangent, so the radius with the tangent is 90 degrees,
OQD = 90 degrees,
and the angle β = OQP = 90 – α.
Then from the triangle POQ, where PO = OQ = r radii, PQ = 7.
By the cosine theorem:
r ^ 2 = 7 ^ 2 + r ^ 2 – 2 * 7 * r * cosγ = 49 + r ^ 2 – 14 * r * cosγ, angle γ = angle POQ = 180 – 2 * β.
14 * r * cosγ = 49.
r = 7 / (2 * cosγ) = 7 / (2 * cos (180 – 2 * (90 – arcos (5 / (4√2))) = 7 / (2 * cos (arcos (5 / (4√ 2))) = 7 / (2 * (5 / (4√2)) = 7 / (5 / (2√2)) = (14√2) / 5.
The radius of the circle is r = (14√2) / 5 = 2.8 * √2.