Q76 / 160 · WBJEE 2010 · Mathematics
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
- Correct answerA$x^2+y^2-2 y-6=0$
- B$x^2+y^2+2 y-6=0$
- C$2\left(x^2+y^2\right)+2 y-6=0$
- D$3\left(x^2+y^2\right)+2 y-6=0$
Solution
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 $x+y-1=0$
$$
\begin{aligned}
& 3-\frac{\lambda}{2}+4-\frac{\lambda}{2}-1=0, \lambda=6 \\
& x^2+y^2-6 x-8 y+6 x+6 y-6=0 ; \quad x^2+y^2-2 y-6=0
\end{aligned}
$$
Hence, the correct answer is option: 'A'.