WBJEE · Maths · Circle
The straight line \(x+y-1=0\) meets the circle \(x^2+y^2-6 x-8 y=0\) at \(A\) and \(B\). Then the equation of the circle of which \(A B\) is a diameter is
- A \(x^2+y^2-2 y-6=0\)
- B \(x^2+y^2+2 y-6=0\)
- C \(2\left(x^2+y^2\right)+2 y-6=0\)
- D \(3\left(x^2+y^2\right)+2 y-6=0\)
Answer & Solution
Correct Answer
(A) \(x^2+y^2-2 y-6=0\)
Step-by-step Solution
Detailed explanation
Hints: \(x^2+y^2-6 x-8 y+\lambda(x+y-1)=0\) Centre \(=\left(3-\frac{\lambda}{2} \cdot 4-\frac{\lambda}{2}\right)\) Lie on \(x+y-1=0\)…
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