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Question

# The equation of circle passing through the origin and cutting off equal intercepts of 2 units on the lines √3y2−√3x2−2xy=0 is/are

A
(x+3)(x1)+(y3)(y1)=0
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B
(x+1)(x+3)+(y+3)(y1)=0
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C
(x3)(x+1)+(y+1)(y+3)=0
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D
(x3)(x1)+(y+1)(y3)=0
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Solution

## The correct options are A (x+√3)(x−1)+(y−√3)(y−1)=0 B (x+1)(x+√3)+(y+√3)(y−1)=0 C (x−√3)(x+1)+(y+1)(y+√3)=0 D (x−√3)(x−1)+(y+1)(y−√3)=0 Given : √3y2−√3x2−2xy=0 ⇒(√3y+x)(y−√3x)=0 ⇒y−√3x=0 ⇒y=√3x ⋯(1) and √3y+x=0 ⇒y=−x√3 ⋯(2) Slope of line (1), tanθ1=√3 ⇒θ1=60∘ ∴A≡(2cos60∘,2sin60∘) or A≡(1,√3) Slope of line (2), tanθ2=−1√3 ⇒θ2=150∘ ∴B≡(2cos150∘,2sin150∘) or B≡(−√3,1) Similarly, C≡(2cos240∘,2sin240∘) or C≡(−1,−√3) and D≡(2cos(−30)∘,2sin(−30)∘) or D≡(√3,−1) Since both these lines are perpendicular to each other so AB,BC,CD and AD will be diameter of the required circles. Therefore using diametric form for equation of circles, with AB as diameter : (x+√3)(x−1)+(y−√3)(y−1)=0 with BC as diameter : (x+1)(x+√3)+(y+√3)(y−1)=0 with CD as diameter : (x−√3)(x+1)+(y+1)(y+√3)=0 with AD as diameter : (x−√3)(x−1)+(y+1)(y−√3)=0

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