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Question

# Let S be the circle in the xy-plane defined by the equation x2+y2=4. Let P be a point on the circle S with both coordinates being positive. Let the tangent to S at P intersect the coordinate axes at the points M and N. Then, the mid-point of the line segment MN must lie on the curve

A
(x+y)2=3xy
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B
x2/3+y2/3=24/3
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C
x2+y2=2xy
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D
x2+y2=x2y2
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Solution

## The correct option is D x2+y2=x2y2Given circle is x2+y2=4 Assuming P=(2cosθ,2sinθ) Let the midpoint of MN=(h,k) Tangent at P, 2cosθ⋅x+2sinθ⋅y−4=0⇒xcosθ+ysinθ−2=0 So, the coordinates of M and N are, M=(2cosθ,0)N=(0,2sinθ) Now, the midpoint is (h,k)=(1cosθ,1sinθ)⇒1h2+1k2=1 Hence, the midpoint of MN lies on the curve x2+y2=x2y2

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