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(A,D) $(1)$ The circle is $x^2+y^2=4$. The point $P_0(1,1)$ lies inside the circle.For chord $E_1E_2$ parallel to the $x$-axis,$y=1$. Substituting int

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(A,D) $(1)$ The circle is $x^2+y^2=4$. The point $P_0(1,1)$ lies inside the circle.For chord $E_1E_2$ parallel to the $x$-axis,$y=1$. Substituting into the circle equation: $x^2+1=4 \implies x^2=3 \implies x=\pm\sqrt{3}$. Thus $E_1(-\sqrt{3}, 1)$ and $E_2(\sqrt{3}, 1)$.The tangent at $(x_1, y_1)$ is $xx_1+yy_1=4$. Tangents at $E_1, E_2$ are $-x\sqrt{3}+y=4$ and $x\sqrt{3}+y=4$. Solving these gives $E_3(0, 4)$.For chord $F_1F_2$ parallel to the $y$-axis,$x=1$. Substituting into the circle equation: $1+y^2=4 \implies y^2=3 \implies y=\pm\sqrt{3}$. Thus $F_1(1, \sqrt{3})$ and $F_2(1, -\sqrt{3})$.Tangents at $F_1, F_2$ are $x+y\sqrt{3}=4$ and $x-y\sqrt{3}=4$. Solving these gives $F_3(4, 0)$.For chord $G_1G_2$ with slope $-1$ passing through $(1,1)$,the line is $y-1=-1(x-1) \implies x+y=2$. Intersection with $x^2+y^2=4$: $x^2+(2-x)^2=4 \implies 2x^2-4x=0 \implies x=0, 2$. Thus $G_1(0, 2)$ and $G_2(2, 0)$.Tangents at $G_1(0, 2)$ is $y=2$. Tangent at $G_2(2, 0)$ is $x=2$. Intersection $G_3(2, 2)$.Points $E_3(0, 4), F_3(4, 0), G_3(2, 2)$ all satisfy $x+y=4$. Correct option is $(A)$.$(2)$ Let $P(2\cos	heta, 2\sin	heta)$. The tangent is $x\cos	heta+y\sin	heta=2$. Intercepts are $M(2/\cos	heta, 0)$ and $N(0, 2/\sin	heta)$.Mid-point $(h, k) = (1/\cos	heta, 1/\sin	heta)$.Thus $\cos	heta=1/h$ and $\sin	heta=1/k$. Since $\cos^2	heta+\sin^2	heta=1$,we have $1/h^2+1/k^2=1 \implies h^2+k^2=h^2k^2$.Locus is $x^2+y^2=x^2y^2$. Correct option is $(D)$.

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