Paragraph: Let S be the circle in the x y – plane defined by the equation x 2   +   y 2…

Paragraph: Let S be the circle in the xy – plane defined by the equation x2 + y2 = 4.

Question : Let E1 E2 and F1 F2 be the chords of S passing through the point P0 (1,1) and parallel to the x – axis and the y – axis, respectively. Let G1 G2 be the chord of S passing through P0 and having slope −1 . Let the tangents to S at E1 and E2 meet at  E3, the 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
  1. x+y=4
  2. x-42+y-42=16
  3. x-4y-4=4
  4. xy=4

Solution

Co - ordinates of E1 and E2 are obtained by solving y=1 and x2+y2=4 ∴  E1-3, 1 and E23, 1 Co - ordinates of F1 and F2 are obtained by solving x=1 and x2+y2=4 F11, 3 and F21,-3 Tangent at E1: -3x+y=4 Tangent at E2: -3x+y=4 ∴E30, 4 Tangent at F1:x+3y=4 Tangent at F2:x-3y=4 ∴  F34, 0 And similarly G32, 2 0, 4,4, 0and 2, 2 lies on x+y=4 Alternate solution 2: The required curve will be the polar of the pole P 0 ( 1,1 ) w.rt. the circle S Hence its equation is T=0 ⇒x.1+y.1=4 ⇒x+y=4 Alternate solution 3: Let Q( h,k ) is a general point on the curve containing points E 3 , F 3   &   G 3 and a pair of tangents are drawn to the circle S from Q then the equation of the chord of contact will be T=xh+yk−4=0 since the chord of contact passes through P 0 ( 1,1 ) hence 1.h+1.k−4=0 ⇒ The point Q( h,k ) lies on the line 1.x+1.y−4=0 ⇒x+y−4=0

Asked in: JEE Advanced 2018 (Paper 1)

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