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Question

Circle x2+y2=5 and x2+y23x+y=0 intersect at point A & B. Lines OA and OB are drawn through origin. Then

A
Angle between OA and OB is 90o
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B
Length of AB is 10 units
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C
Area of ΔAOB is 5 sq. units
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D
Tangents at A and B to x2+y2=5 intersect at (3,1)
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Solution

The correct options are
A Length of AB is 10 units
B Angle between OA and OB is 90o
D Tangents at A and B to x2+y2=5 intersect at (3,1)
(A) Solving two equations of circle we get A=(2,1) and B=(1,2).
So equation of AB is 3xy5=0.
On homogenising AB with x2+y2=5,
Combined equation of OA & OB is
x2+y25(3xy5)2=0
or 4x24y26xy=0
angle between OA & OB is 90o

(B) LAB=2×AB=2(OA)2(OM)2
LAB=2 5(510)2=10
(C) Area of ΔAOB=12(AB)(OM)
=1210.510=52
(D) Let tangents at A & B intersect at (a,b)
AB is ax+by5=0 , A=(1,2),B=(2,1)
we get intersection point as(3,1)
Ans: A,B,D

336492_290799_ans_a8c5540c991d4fdfaeceadfd304c7709.png

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