The straight line x+y−1=0 meets the circle x2+y2−6x−8y=0 at A and B. Then the equation of the circle of which AB is a diameter is
A
x2+y2−2y−6=0
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B
x2+y2+2y−6=0
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C
2(x2+y2)+2y−6=0
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D
3(x2+y2)+2y−6=0
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Solution
The correct option is Dx2+y2−2y−6=0 x2+y2−6x−8y+λ(x+y−1)=0 ∴ Centre =(3−λ2,4−λ2) lies on x+y−1=0 So, the centre satisfy the equation x+y−1=0 ∴(3−λ2)+(4−λ2)−1=0 ⇒6−λ=0⇒λ=6 Hence, required equation is given by x2+y2−6x−8y+6x+6y−6=0 ⇒x2+y2−2y−6=0