0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
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 ā1. 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

A
x+y=4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
(x4)2+(y4)2=16
No worries! Weāve got your back. Try BYJUāS free classes today!
C
(x4)(y4)=4
No worries! Weāve got your back. Try BYJUāS free classes today!
D
xy=4
No worries! Weāve got your back. Try BYJUāS free classes today!
Open in App
Solution

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

Suggest Corrections
2
Join BYJU'S Learning Program
Related Videos
Chord of a Circle
MATHEMATICS
Watch in App
Join BYJU'S Learning Program