The sides a, b, c of a right triangle, where c is the hypotenuse, are circumscribing a circle. Prove that the radius r of the circle . . .

Question : The sides a, b, c of a right triangle, where c is the hypotenuse, are circumscribing a circle. Prove that the radius r of the circle is given by r=(a+b-c)/2

Doubt by Muskan

Solution : 

We know, length of the tangents drawn from an external point to a circle are equal. 

CP = CR
BQ = BR
AP = AQ

Also, 
OP = OR [Radii of the same circle]
OP⊥CA [Tangent is perpendicular to radius through the point of contact]
OR⊥BC
OQ⊥AB

∴ PORC is a square of side r.
CP=CR = r

BR=BQ = a-r
AP=AQ = b-r

AB = AQ+BQ
AB = AP+BR
c = b-r + a-r
c=a+b-2r
2r=a+b-c
r=(a+b-c)/2