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Question

Let 3x−y−3=0 is a diameter of the circle x2+y2−4x−6y+4=0. If L1 is the chord which is bisected by the given diameter line, then which of the following is/are true?

A
When L1 is the largest possible chord of the circle, its possible equation is x+3y11=0
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B
When L1 is the largest possible chord of the circle, its possible equation is x3y+8=0
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C
When L1 is at a distance of 1 from the centre, its equation is x+3y11+10=0
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D
When L1 is at a distance of 1 from the centre, its equation is x+3y1110=0
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Solution

The correct option is D When L1 is at a distance of 1 from the centre, its equation is x+3y1110=0
Given circle is x2+y24x6y+4=0
C=(2,3), r=3
Diameter is 3xy3=0
The largest chord is diameter and two diameter always bisect each other.
Checking the equations
x+3y11=0 and x3y+8=0
Diameter passes through the centre (2,3), so
2+911=029+8=10
Therefore x+3y11=0 can be equation of diameter.

The chord which is bisected by the given diameter is perpendicular to it, so the equation of chord is
x+3y+λ=0
Distance from centre is 1, so
|11+λ|10=1λ=11±10
Hence, the required equation of chord are
x+3y11+10=0x+3y1110=0

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