How to find the distance (in a right triangle) between the centers of the inscribed circle?

Legend:
∠C = 90 °;
AB = c;
BC = a;
AC = b;
R is the radius of the circumscribed circle;
r is the radius of the inscribed circle.

Inscribed and circumcircle properties
Let points M and N be the centers of the inscribed and circumscribed circles.

Since in triangle ABC the hypotenuse AB is the diameter of the circumscribed circle, point N lies in the middle of AB.

An MQCL quadrilateral is a rectangle because:

∠C = 90 °;

∠MQC = ∠MQC = 90 °, because the radius is perpendicular to the tangent;

∠QML = 90 °, because the sum of the angles of the quadrilateral is 360 °,

but since the adjacent sides are equal: MQ = ML = r, then it is also a square.

Calculating the length of a segment NP
Line segments BP and BL are tangent to the incircle. Therefore, the distance from point B to points of tangency P and L are equal:

BP = BL = BC – LC = a – r.

We calculate the length of the segment NP:

NP = BP – BN;

NP = a – r – R;

NP = a – (R + r).

Similarly, we obtain for the tangents AP and AQ drawn from point A:

AP = AQ = AC – AQ = b – r;

NP = AN – AP;

NP = R – (b – r);

NP = (R + r) – b.

From these two equalities it follows:

a + b = 2 * (R + r);

NP = 1/2 * (a – b).

Calculating the distance between the centers of the circles
We apply the Pythagorean theorem to the right-angled triangle MPN to determine the unknown hypotenuse MN, which is the distance between the centers of the inscribed and circumscribed circles:

MN² = NP² + MP²;
MN² = 1/4 * (a – b) ² + r²;
N = √ {r² + ((a – b) / 2) ²}.
Let’s check for an isosceles triangle (a = b):

MN = r.

Answer: √ {r ^ 2 + ((a – b) / 2) ^ 2}.