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Question

Let S be the circle in the xy-plane defined by the equation x2+y2=4.

Let E1E2 and F1F2 be the chords of S passing through the point P0(1,1) and parallel to the x-axis and the y-axis, respectively. Let G1G2 be the chord of S passing through P0 and having slope ā1. Let the tangents to S at E1 and E2 meet at E3, the tangents to S at F1 and F2 meet at F3, and the tangents to S at G1 and G2 meet at G3. Then, the points E3,F3, and G3 lie on the curve

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Solution

The correct option is **A** x+y=4

The equation of circle: x2+y2=4

Coordinates of points,

E1=(√3,1)E2=(−√3,1)

Tangents at E1 and E2,

√3x+y=4..........(i)−√3x+y=4......(ii)

Using equation (i) and (ii),

E3=(0,4)

Similarly,

F1=(1,√3)F2=(1,−√3)

Tangents at F1 and F2,

x+√3y=4..........(iii)x−√3y=4......(iv)

Using equation (iii) and (iv),

F3=(4,0)

Now for G3,

G1=(2,0)G2=(0,2)

Tangents at G1 and G2,

2x=4..........(v)2y=4......(vi)

Using equation (v) and (vi),

G3=(2,2)

Hence the points E3=(0,4),F3=(4,0),G3=(2,2) lies on

x+y=4

The equation of circle: x2+y2=4

Coordinates of points,

E1=(√3,1)E2=(−√3,1)

Tangents at E1 and E2,

√3x+y=4..........(i)−√3x+y=4......(ii)

Using equation (i) and (ii),

E3=(0,4)

Similarly,

F1=(1,√3)F2=(1,−√3)

Tangents at F1 and F2,

x+√3y=4..........(iii)x−√3y=4......(iv)

Using equation (iii) and (iv),

F3=(4,0)

Now for G3,

G1=(2,0)G2=(0,2)

Tangents at G1 and G2,

2x=4..........(v)2y=4......(vi)

Using equation (v) and (vi),

G3=(2,2)

Hence the points E3=(0,4),F3=(4,0),G3=(2,2) lies on

x+y=4

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