The curve xy = c, (c 0),and the circle x^2 | vedclass

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(B) Given the curve $xy = c$ and the circle $x^2 + y^2 = 1$.Since the curves touch each other,they share a common tangent at the points of contact.By

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$2$

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(B) Given the curve $xy = c$ and the circle $x^2 + y^2 = 1$.Since the curves touch each other,they share a common tangent at the points of contact.By symmetry,if $P(x, y)$ is a point of contact,then $Q(-x, -y)$ is also a point of contact.Both points $P$ and $Q$ lie on the circle $x^2 + y^2 = 1$.The line segment $PQ$ passes through the origin $O(0, 0)$ because the points are symmetric with respect to the origin.Since $P$ and $Q$ are on the circle and the line $PQ$ passes through the center of the circle,$PQ$ is a diameter of the circle.The radius of the circle $x^2 + y^2 = 1$ is $r = 1$.Therefore,the distance between the points of contact $P$ and $Q$ is the diameter of the circle,which is $2r = 2(1) = 2$.

The curve $xy = c, (c > 0)$,and the circle $x^2 + y^2 = 1$ touch at two points. Then the distance between the points of contact is

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