In a right-angled triangle ABC with a right angle C and leg BC = 9, the radius of the inscribed circle is 3.

In a right-angled triangle ABC with a right angle C and leg BC = 9, the radius of the inscribed circle is 3. Find: a) sides AB and AC; b) the distance between the centers of the inscribed and circumscribed circles.

a) 1. KO = OH = 3 (radii of a circle centered at point O).
Since BC = 9, then BH = BC – CH = 9 – 3 = 6.
2. Consider the circle and angle HBN.
This property is fulfilled: if the circle of the sides touches the sides of the corner, then the segments of the tangents are equal to each other. That is, HB = NB = 6.
3. Similarly to point 2, the segments AK and AN are equal.
4. Let AK = AN = x.
Then, by the Pythagorean theorem, (x + 3) ^ 2 + 9 ^ 2 = (x + 6) ^ 2.
Solving this equation, we get: x = 9.
AB = x + 6 = 9 + 6 = 15.
AC = x + 3 = 9 + 3 = 12.
Answer: AB = 15; AC = 12.
b) 1. Let G be the center of the circumscribed circle.
Let’s use the property: The center of a circle circumscribed about a right-angled triangle is the midpoint of the hypotenuse. That is, AG = GB = AB / 2 = 7.5.
GN = GB – NB = 7.5 – 6 = 1.5.
2. Find the distance between the centers of the inscribed and circumscribed circles:
OG = (GN ^ 2 + ON ^ 2) ^ (1/2);
OG = (1.5 ^ 2 + 3 ^ 2) ^ (1/2);
OG = (11.25) ^ (1/2).
Answer: OG = (11.25) ^ (1/2).