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Question

Tangent to the circle x2+y2 = 4 at any point on it in the first quadrant makes intercepts OA and OB on x and y axes respectively, O being the centre of circle. Find the minimum value of (OA + OB).

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Solution

Given circle x2+y2=4 ....(i) Let P (α,β) α2+β2 ....(ii)

Also 2x+2ydydx=0 dydx=xy dydxat P=αβ=mr

Equation of tangent : yβ=αβ(xa) αx+βy=a2+β2

By (ii), αx+βy=4 x4α+y4β=1 xintercept=4α=OA,yintercept=4β=OB

Let OA+OB=s=4α+4β s=4α+44α2 dsdα=4α2+4α(4α2)32

d2sdα2=8α3+(4α2)3244α×32(4α2)12(2α)(4α2)3 d2sdα2=8α3+8(α2+2)(4α2)52

For local maxima and/or minima, dsdα=4α2+4α(4α2)32=0 α3=(4α2)32

α6=(4α2)3=6448α2+12α4α6 m36m2+24m32=0, where m = α2

(m2)=0 or (m24m+16)=0 m=2,m24m+16 0 m ϵ R

α=2. [ P lies in 1st quadrant] d2sdα2at α=2=822+8(2+2)(42)52=62>0

So, s is minimum at α=2.

Therefore, the minimum value of s =42+442=42.


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