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If $x-y+1=0$ meets the circle $x^2+y^2+y-1=0$ at $A$ and $B$, then the equation of the circle with $A B$ as diameter is
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Verified Answer
The correct answer is:
$2\left(x^2+y^2\right)+3 x-y+1=0$
Given that circle, $S=x^2+y^2+y-1=0$
Line, $\mathrm{L}: x-y+1=0$
Equation of the circle passing through the intersection of line and circle is given by
$\begin{aligned}
& \mathrm{S}+\lambda \mathrm{L}=0 \\
& \left(x^2+y^2+y-1\right)+\lambda(x-y+1)=0 \\
& x^2+y^2+\lambda x+(1-\lambda) y+\lambda-1=0
\end{aligned}$
Centre of above circle $=\left(\frac{-\lambda}{2}, \frac{\lambda-1}{2}\right)$
Since, centre lies on $x-y+1=0$
$\therefore \quad \frac{-\lambda}{2}-\left(\frac{\lambda-1}{2}\right)+1=0 \Rightarrow \lambda=\frac{3}{2}$
Now, the required equation of circle
$\begin{aligned}
& x^2+y^2+y-1+\frac{3}{2}(x-y+1)=0 \\
& \therefore \quad 2\left(x^2+y^2\right)+3 x-y+1=0
\end{aligned}$
Line, $\mathrm{L}: x-y+1=0$
Equation of the circle passing through the intersection of line and circle is given by
$\begin{aligned}
& \mathrm{S}+\lambda \mathrm{L}=0 \\
& \left(x^2+y^2+y-1\right)+\lambda(x-y+1)=0 \\
& x^2+y^2+\lambda x+(1-\lambda) y+\lambda-1=0
\end{aligned}$
Centre of above circle $=\left(\frac{-\lambda}{2}, \frac{\lambda-1}{2}\right)$
Since, centre lies on $x-y+1=0$
$\therefore \quad \frac{-\lambda}{2}-\left(\frac{\lambda-1}{2}\right)+1=0 \Rightarrow \lambda=\frac{3}{2}$
Now, the required equation of circle
$\begin{aligned}
& x^2+y^2+y-1+\frac{3}{2}(x-y+1)=0 \\
& \therefore \quad 2\left(x^2+y^2\right)+3 x-y+1=0
\end{aligned}$
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