Let S be the circle in the xy-plane defined by the equation $x^{2} + y^{2} = 4$ . Let $E_{1} E_{2}$ and $F_{1} F_{2}$ be the chord of S passing through the point $P_{0} (1, 1)$ and parallel to the x-axis and the y-axis, respectively. Let $G_{1} G_{2}$ be the chord of S passing through $P_{0}$ and having slope -1. Let the tangents to S at $E_{1}$ and $E_{2}$ meet at $E_{3}$ , the tangents of S at $F_{1}$ and $F_{2}$ meet at $F_{3}$ , and the tangents to S at $G_{1}$ and $G_{2}$ meet at $G_{3}$ . Then, the points $E_{3}, F_{3} a n d G_{3}$ lie on the curve

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$x + y = 4$

${(x - 4)}^{2} + {(y - 4)}^{2} = 16$

$(x - 4) (y - 4) = 4$

$x y = 4$

answer is A.

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Detailed Solution

Co-ordinates of $E_{1}$ and $E_{2}$ are obtained bysolving $y = 1$ and $x^{2} + y^{2} = 4$
$∴ E_{1} (- \sqrt{3}, 1)$ and $E_{2} (\sqrt{3}, 1)$
Co-ordinates of $F_{1}$ and $F_{2}$ are obtained by solving
$x = 1 a n d x^{2} + y^{2} = 4$

$F_{1} (1, \sqrt{3}) a n d F_{2} (1, - \sqrt{3})$

Tangent at $E_{1} : - \sqrt{3} x + y = 4$
Tangent at $E_{2} : \sqrt{3} x + y = 4$
$∴ E_{5} (0, 4)$
Tangent at : $F_{1} : x + \sqrt{3} y = 4$
Tangent at $F_{2} : x - \sqrt{3} y = 4$
$∴ F_{5} (4, 0)$
And similarly $G_{5} (2, 2)$
$(0, 4), (4, 0)$ and $(2, 2)$ lies on $x + y = 4$