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If $\mathrm{A}$ and $\mathrm{B}$ are the points of intersection of the circles $x^2+y^2-4 x+6 y-3=0$ and $x^2+y^2+2 x-2 y-2=0$ then the distance between $\mathrm{A}$ and $\mathrm{B}$ is
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Verified Answer
The correct answer is:
$\frac{\sqrt{231}}{5}$
$\begin{aligned} & \text {} S_1: x^2+y^2-4 x+6 y-3=0.....(i) \\ & S_2: x^2+y^2+2 x-2 y-2=0.....(ii)\end{aligned}$
$$
\begin{aligned}
& S_1:(x-2)^2+(y+3)^2=4^2 \\
& S_2:(x+1)^2+(y-1)^2=2^2
\end{aligned}
$$
Here, $\mathrm{OA}=2, \mathrm{O}^{\prime} \mathrm{A}=4, \mathrm{O} \equiv(-1,1) \& \mathrm{O}^{\prime} \equiv(2,-3)$
$$
\therefore \mathrm{OO}^{\prime}=5
$$
Let $O P=x \Rightarrow O^{\prime} P=5-x$ \& $A P=d$
In $\triangle \mathrm{AOP}, 2^2=\mathrm{d}^2+\mathrm{x}^2$.....(1)
In $\triangle \mathrm{APO}, 4^2=\mathrm{d}^2+(5-\mathrm{x})^2$.....(2)
From eq ${ }^{\mathrm{n}}(1) \&$ (2)
$$
=\mathrm{x}=\frac{13}{10}, \mathrm{~d}=\sqrt{\frac{231}{100}}=\frac{\sqrt{231}}{10}
$$
Now, $\mathrm{AB}=2 \mathrm{AP}=2 \times \mathrm{d}=2 \times \frac{\sqrt{231}}{10}=\frac{\sqrt{231}}{5}$
$$
\begin{aligned}
& S_1:(x-2)^2+(y+3)^2=4^2 \\
& S_2:(x+1)^2+(y-1)^2=2^2
\end{aligned}
$$
Here, $\mathrm{OA}=2, \mathrm{O}^{\prime} \mathrm{A}=4, \mathrm{O} \equiv(-1,1) \& \mathrm{O}^{\prime} \equiv(2,-3)$
$$
\therefore \mathrm{OO}^{\prime}=5
$$
Let $O P=x \Rightarrow O^{\prime} P=5-x$ \& $A P=d$
In $\triangle \mathrm{AOP}, 2^2=\mathrm{d}^2+\mathrm{x}^2$.....(1)
In $\triangle \mathrm{APO}, 4^2=\mathrm{d}^2+(5-\mathrm{x})^2$.....(2)
From eq ${ }^{\mathrm{n}}(1) \&$ (2)
$$
=\mathrm{x}=\frac{13}{10}, \mathrm{~d}=\sqrt{\frac{231}{100}}=\frac{\sqrt{231}}{10}
$$
Now, $\mathrm{AB}=2 \mathrm{AP}=2 \times \mathrm{d}=2 \times \frac{\sqrt{231}}{10}=\frac{\sqrt{231}}{5}$
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