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 y−7x−3=−23 or 2x+3y−27=0 The family of circles is (x−3)(x−6)+(y−7)(y−5)+λ(2x+3y−27)=0 or x2+y2−9x−12y+53+λ(2x+3y−27)=0 or x2+y2+(2λ−9)x+(3λ−12)y+53−27λ=0 The chord along which this cuts the circle x2+y2−4x−6y−3=0 is S−S′=0 (2λ−5)x+(3λ−6)y+56−27λ=0 or λ(2x+3y−27)−(5x+6y−56)=0 These lines are concurrent at the point given by 2x+3y−27=0 and 5x+6y−56=0 Using Rule of Cross multiplication, we get ∴x3(−56)+6×27=y−5×27+2×56=12×6−5×3 Solving x3(−56)+6×27=12×6−5×3 and y−5×27+2×56=12×6−5×3 we get x=2,y=233