Equation of Family of Circles Touching a Line and Passing through a Given Point on the Line
A circle touc...
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
ba√a2+b2 units
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B
b2a√a2+b2 units
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C
b4a√a2+b2 units
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D
None of these
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Solution
The correct option is Cb4a√a2+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 (x−a2)2+(y−b2)2+λ(xa+yb−1)=0⋯(i) It passes through Q(a2,0) Therefore, λ=b22 On putting the value of λ in Eq. (i), we get (x−a2)2+(y−b2)2+b22(xa+yb−1)=0 ⇒x2+y2−(a−b22a)x−by2+a2−b24=0 Let r be the radius of this circle. Then, r2=14(a−b22a)2+b216−(a2−b24)=b2(a2+b2)16a2 ⇒r=b4a√a2+b2 units
Alternate Solution: Equation of PC will be xb−ya=a2−b22ab⋯(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=2a2−b24a ∴r=√(2a2−b24a−a2)2+b216 ⇒r=b4a√a2+b2 units