Mathematics - Circle Question with Solution | TestHub

Co - ordinates of$E_1 \mathrm{a}\mathrm{n}\mathrm{d} E_2$are obtained by solving$y=1 \mathrm{a}\mathrm{n}\mathrm{d} x^2+y^2=4$$\therefore E_1\left(-\sqrt{3}, 1\right) \mathrm{a}\mathrm{n}\mathrm{d} E_2\left(\sqrt{3}, 1\right)$Co - ordinates of$F_1 \mathrm{a}\mathrm{n}\mathrm{d} F_2$are obtained by solving$x=1 \mathrm{a}\mathrm{n}\mathrm{d} x^2+y^2=4$$F_1\left(1, \sqrt{3}\right) \mathrm{a}\mathrm{n}\mathrm{d} F_2\left(1,-\sqrt{3}\right)$Tangent at$E_1: -\sqrt{3}x+y=4$Tangent at$E_2: -\sqrt{3}x+y=4$$\therefore E_3\left(0, 4\right)$Tangent at$F_1:x+\sqrt{3}y=4$Tangent at$F_2:x-\sqrt{3}y=4$$\therefore F_3\left(4, 0\right)$And similarly$G_3\left(2, 2\right)$$\left(0, 4\right),\left(4, 0\right)\mathrm{a}\mathrm{n}\mathrm{d} \left(2, 2\right)\mathrm{ }\mathrm{l}\mathrm{i}\mathrm{e}\mathrm{s} \mathrm{o}\mathrm{n} x+y=4$Alternate solution 2: The required curve will be the polar of the pole$P_0\left(1,1\right)$w.rt. the circle S Hence its equation is$T=0$$\Rightarrow x.1+y.1=4$$\Rightarrow x+y=4$Alternate solution 3: Let$Q\left(h,k\right)$is a general point on the curve containing points$E_3,F_3\&G_3$and a pair of tangents are drawn to the circle S from Q then the equation of the chord of contact will be$T=xh+yk-4=0$since the chord of contact passes through$P_0\left(1,1\right)$hence$1.h+1.k-4=0$$\Rightarrow$The point$Q\left(h,k\right)$lies on the line$1.x+1.y-4=0$$\Rightarrow x+y-4=0$