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Equation of a Chord Joining Two Points with Circle in Parametric Form

Let S be the ...

Question

Let $S$ be the circle in the $x y$ -plane defined by the equation $x^{2} + y^{2} = 4$ .

Let $E_{1} E_{2}$ and $F_{1} F_{2}$ be the chords of $S$ passing through the point $P_{0} (1, 1)$ and parallel to the $x$ -axis and the $y$ -axis, respectively. Let $G_{1} G_{2}$ be the chord of $S$ passing through $P_{0}$ and having slope $- 1$ . Let the tangents to $S$ at $E_{1}$ and $E_{2}$ meet at $E_{3}$ , the tangents to $S$ at $F_{1}$ and $F_{2}$ meet at $F_{3}$ , and the tangents to $S$ at $G_{1}$ and $G_{2}$ meet at $G_{3}$ . Then, the points $E_{3}, F_{3}$ , and $G_{3}$ lie on the curve

x + y = 4

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(x - 4)^{2} + (y - 4)^{2} = 16

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(x - 4) (y - 4) = 4

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x y = 4

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Solution

The correct option is A $x + y = 4$
The equation of circle: $x^{2} + y^{2} = 4$

Coordinates of points,
$E_{1} = (\sqrt{3}, 1) E_{2} = (- \sqrt{3}, 1)$
Tangents at $E_{1}$ and $E_{2}$ ,
$\sqrt{3} x + y = 4.......... (i) - \sqrt{3} x + y = 4 . . . . . . (i i)$
Using equation (i) and (ii),
$E_{3} = (0, 4)$
Similarly,
$F_{1} = (1, \sqrt{3}) F_{2} = (1, - \sqrt{3})$
Tangents at $F_{1}$ and $F_{2}$ ,
$x + \sqrt{3} y = 4.......... (i i i) x - \sqrt{3} y = 4 . . . . . . (i v)$
Using equation (iii) and (iv),
$F_{3} = (4, 0)$
Now for $G_{3}$ ,
$G_{1} = (2, 0) G_{2} = (0, 2)$
Tangents at $G_{1}$ and $G_{2}$ ,
$2 x = 4.......... (v) 2 y = 4 . . . . . . (v i)$
Using equation (v) and (vi),
$G_{3} = (2, 2)$
Hence the points $E_{3} = (0, 4), F_{3} = (4, 0), G_{3} = (2, 2)$ lies on
$x + y = 4$

Suggest Corrections

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