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Question

If OA and OB are equal perpendicular chords of the circles x2+y2−2x+4y=0, then equation of OA and OB are where O is origin.

A
3x+y=0 and 3xy=0
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B
3x+y=0 or 3yx=0
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C
x+3y=0 and y3x=0
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D
x+y=0 or xy=0
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Solution

The correct option is B x+3y=0 and y3x=0
Let the equation of the chord OA of a circle
x2+y22x+4y=0 ......... (i)
be y=mx ........ (ii)
Solving (i) and (ii), we get
x2+m2x22x+4mx=0
(1+m2)x2(24m)x=0
x=0 and x=24m1+m2
Hence, the points of intersection are
(0,0) and A(24m1+m2,m(24m)1+m2)
OA2=(24m1+m2)2(1+m2)
=(24m)21+m2
Since OAB is an isosceles right-angled triangle OA2=12AB2
AB is a diameter of the given circle
OA2=10

(24m)21+m2=10

416+16m2=10(1+m)2

3m28m3=0
m=3 or 13
Hence, the required equations are y=3x or x+3y=0

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