In a right-angled triangle ABC, leg AC = 12, BC = 5. Find the radius of the circle that passes through the ends of the hypotenuse of the triangle and touches BC.
Let’s finish the right-angled triangle ABC to the rectangle AСВM.
AM = BC = 5 cm, CM = AC = 12 cm.
The segment BC is tangent to the circle, MB is perpendicular to the tangent, which means that BM passes through the point O – the center of the circle.
Extend ВM to the intersection with the circle at point K.
The AВK triangle is rectangular, since ВK is the diameter of the circle, then AM, the height, is drawn from the vertex of the right angle to the hypotenuse.
AM ^ 2 = MK * ВM.
25 = 12 * MK.
MK = 25/12.
ВK = ВM + MK = 12 + 25/12 = 169/12.
Then R = ВO = ВK / 2 = 169/24 = 7 (1/24) cm.
Answer: The radius of the circle is 7 (1/24) cm.
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