JEE Advanced 2018 — Paper 1 — Question 45

The circle is $x^2 + y^2 = 4$. Chord $E_1E_2$ is horizontal through $P_0(1,1)$, so its equation is $y = 1$.

Substituting into the circle gives $x^2 + 1 = 4 \Rightarrow x = \pm \sqrt{3}$.

Thus $E_1(-\sqrt{3},1), E_2(\sqrt{3},1)$. Tangents at $E_1$ and $E_2$ are $-\sqrt{3}x + y = 4$ and $\sqrt{3}x + y = 4$. Solving gives $E_3(0,4)$. Chord $F_1F_2$ is vertical through $P_0$, so $x = 1$.

Substituting gives $1 + y^2 = 4 \Rightarrow y = \pm \sqrt{3}$.

Thus $F_1(1,-\sqrt{3}), F_2(1,\sqrt{3})$. Tangents at $F_1$ and $F_2$ are $x - \sqrt{3}y = 4$ and $x + \sqrt{3}y = 4$. Solving gives $F_3(4,0)$. Chord $G_1G_2$ has slope $-1$ through $P_0$, so its equation is $y - 1 = -1(x - 1) \Rightarrow y = -x + 2$.

Substituting into the circle: $x^2 + (-x+2)^2 = 4 \Rightarrow 2x^2 - 4x + 4 = 4 \Rightarrow 2x(x-2)=0 \Rightarrow x=0,2$.

Corresponding $y$: $y=2,0$.

Thus $G_1(0,2), G_2(2,0)$. Tangents at $G_1$ and $G_2$ are $0\cdot x + 2y = 4 \Rightarrow y=2$ and $2x + 0\cdot y = 4 \Rightarrow x=2$.

Solving gives $G_3(2,2)$. Points $E_3(0,4), F_3(4,0), G_3(2,2)$ all satisfy $x + y = 4$.

Hence they lie on the line $x + y = 4$.