A circle touches the hypotenuse of a right-angled triangle at its middle point and passes through the mid-point of the shorter side. If a and b(a<b) be the length of the sides, then prove that the radius is b4a√a2+b2.
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Choose the perpendicular sides along axes of co-ordinates so that the hypotenuse is xa+yb=1 which is a tangent at mid-point (a/2,b/2). The equation of the circle by (n1) is (x−a2)2+(y−b2)2Pointcircle+λ(xa+yb−1)Tangent=0 It passes through mid-point Q of OA i.e. (a2,0) ∴b24−λ2=0 or λ=b22 Required circle is (x−a2)2+(y−b2)2+b22(xa+yb−1)=0 or x2+y2+x(−a+b22a)+y(−b+b2)+a2+b24−b22=0 or x2+y2+b2−2a22ax−b2y+a2−b24=0 ∴r2=g2+f2−c =(b2−2a24a)2+(b4)2−(a2−b2)4 =b416a2+a24−b24+b216−a24+b24 =b4+a2b216a2∴r=b4a√a2+b2.