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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
$\Rightarrow x^2+(x+1)^2+x+1-1=0 \quad$ [from Eq. (i)]
$\begin{array}{lc}
\Rightarrow & 2 x^2+3 x+1=0 \\
\Rightarrow & (2 x+1)(x+1)=0 \\
\Rightarrow & x=-\frac{1}{2},-1 \text { and } y=\frac{1}{2}, 0
\end{array}$
$\therefore$ Point of $A\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $B(-1,0)$
These are the end points of a diameter.
$\therefore$ The equation of circle is
$\left(x+\frac{1}{2}\right)(x+1)+\left(y-\frac{1}{2}\right)(y-0)=0$
$\begin{array}{rrrl}\Rightarrow & (2 x+1)(x+1)+(2 y-1) y=0 \\ \Rightarrow & 2 x^2+x+2 x+1+2 y^2-y=0 \\ \Rightarrow & 2\left(x^2+y^2\right)+3 x-y+1=0\end{array}$
$\Rightarrow x^2+(x+1)^2+x+1-1=0 \quad$ [from Eq. (i)]
$\begin{array}{lc}
\Rightarrow & 2 x^2+3 x+1=0 \\
\Rightarrow & (2 x+1)(x+1)=0 \\
\Rightarrow & x=-\frac{1}{2},-1 \text { and } y=\frac{1}{2}, 0
\end{array}$
$\therefore$ Point of $A\left(-\frac{1}{2}, \frac{1}{2}\right)$ and $B(-1,0)$
These are the end points of a diameter.
$\therefore$ The equation of circle is
$\left(x+\frac{1}{2}\right)(x+1)+\left(y-\frac{1}{2}\right)(y-0)=0$
$\begin{array}{rrrl}\Rightarrow & (2 x+1)(x+1)+(2 y-1) y=0 \\ \Rightarrow & 2 x^2+x+2 x+1+2 y^2-y=0 \\ \Rightarrow & 2\left(x^2+y^2\right)+3 x-y+1=0\end{array}$
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