JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 15)
Let S be the circle in the XY-plane defined the equation x2 + y2 = 4.
Let E1E2 and F1F2 be the chords of S passing through the point P0 (1, 1) and parallel to the X-axis and the Y-axis, respectively. Let G1G2 be the chord of S passing through P0 and having slope$$-$$1. Let the tangents to S at E1 and E2 meet at E3, then tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3 and G3 lie on the curve
Let E1E2 and F1F2 be the chords of S passing through the point P0 (1, 1) and parallel to the X-axis and the Y-axis, respectively. Let G1G2 be the chord of S passing through P0 and having slope$$-$$1. Let the tangents to S at E1 and E2 meet at E3, then tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3, F3 and G3 lie on the curve
x + y = 4
(x $$-$$ 4)2 + (y $$-$$ 4)2 = 16
(x $$-$$ 4)(y $$-$$ 4) = 4
xy = 4
Explanation
Equation of tangent at $${E_1}( - \sqrt 3 ,1)$$ is
$$ - \sqrt 3 $$x + y = 4 and at $${E_2}(\sqrt 3 ,1)$$ is $$\sqrt 3 $$x + y = 4
Intersection point of tangent at E1 and E2 is (0, 4)
$$ \therefore $$ Coordinates of E3 is (0, 4)
Similarly, equation of tangent at
$${F_1}(1, - \sqrt 3 )$$ and $${F_2}(1,\sqrt 3 )$$ are x $$-$$ $$\sqrt 3 $$y = 4 and
x + $$\sqrt 3 $$y = 4, respectively and intersection point
is (4, 0), i.e., F3(4, 0) and equation of tangent at G1(0, 2) and G2(2, 0) are 2y = 4 and 2x = 4, respectively and intersection point
is (2, 2) i.e., G3(2, 2).
Point E3(0, 4), F3(4, 0) and G3(2, 2) satisfies the line x + y = 4.
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