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Question

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 the radius of circle is

A
baa2+b2 units
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B
b2aa2+b2 units
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C
b4aa2+b2 units
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D
None of these
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Solution

The correct option is C b4aa2+b2 units
The equation of the circle touching xa+yb=1 at P (a2,b2) is S+λL=0 Where S=0 is point circle
(xa2)2+(yb2)2+λ(xa+yb1)=0(i)
It passes through Q(a2,0)
Therefore, λ=b22
On putting the value of λ in Eq. (i), we get
(xa2)2+(yb2)2+b22(xa+yb1)=0
x2+y2(ab22a)xby2+a2b24=0
Let r be the radius of this circle.
Then,
r2=14(ab22a)2+b216(a2b24)=b2(a2+b2)16a2
r=b4aa2+b2 units

Alternate Solution:
Equation of PC will be
xbya=a2b22ab(1)
Perpendicular bisector of the chord will always pass through the centre of the circle.
Let the coordinate of C will be
(α,b4)
Putting the value of C in (1)
α=2a2b24a
r=(2a2b24aa2)2+b216
r=b4aa2+b2 units

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