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Question

Consider a family of circles passing through the points (3,7) and (6,5). The chords in which the circle x2+y2−4x−6y−3=0 cuts the family of circles are concurrent at the point

A
(2,233)
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B
(2,233)
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C
(2,233)
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D
(2,233)
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Solution

The correct option is B (2,233)
The line through (3,7) and (6,5) is y7x3=23
or 2x+3y27=0
The family of circles is (x3)(x6)+(y7)(y5)+λ(2x+3y27)=0
or x2+y29x12y+53+λ(2x+3y27)=0
or x2+y2+(2λ9)x+(3λ12)y+5327λ=0
The chord along which this cuts the circle
x2+y24x6y3=0 is SS=0
(2λ5)x+(3λ6)y+5627λ=0
or λ(2x+3y27)(5x+6y56)=0
These lines are concurrent at the point given by
2x+3y27=0 and 5x+6y56=0
Using Rule of Cross multiplication, we get
x3(56)+6×27=y5×27+2×56=12×65×3
Solving x3(56)+6×27=12×65×3 and y5×27+2×56=12×65×3
we get x=2,y=233

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