Paragraph: Let S be the circle in the xy – plane defined by the...

Co - ordinates of ${E}_{1}and{E}_{2}$ are obtained by solving $y = 1and{x}^{2} + {y}^{2} = 4 ∴ {E}_{1}$\left(- \sqrt{3}, 1\right)$and{E}_{2}$\left(\sqrt{3}, 1\right)$Co - ordinates of${F}_{1}and{F}_{2}$are obtained by solving$x = 1and{x}^{2} + {y}^{2} = 4 {F}_{1}\left(1, \sqrt{3}\right)and{F}_{2}\left(1, - \sqrt{3}\right) Tangent at ${E}_{1}: - \sqrt{3}x + y = 4$ Tangent at ${E}_{2}: - \sqrt{3}x + y = 4 ∴ {E}_{3}\left(0, 4\right)$ Tangent at ${F}_{1}: x + \sqrt{3}y = 4$ Tangent at ${F}_{2}: x - \sqrt{3}y = 4 ∴ {F}_{3}$\left(4, 0\right)$And similarly{G}_{3}$\left(2, 2\right) $\left(0, 4\right)$, \left(4, 0\right)$and$\left(2, 2\right)$liesonx + y = 4$ 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 ${E}_{3} , {F}_{3}$\text{and } {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