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If $(3,1)$ and $(-2,4)$ are points on a circle $S$ whose centre lies on the line $x-y+1=0$ then the parametric equations of S are
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Verified Answer
The correct answer is:
$x=-1+\sqrt{17} \cos \theta, y=\sqrt{17} \sin \theta$
Centre lies on $x-y+1=0$
$\therefore \quad$ Centre: $C(\alpha, \alpha+1)$
Let $P(3,1)$ and $Q(-2,4)$ are 2 points
$$
\begin{aligned}
& \therefore \quad C P=C Q \Rightarrow C P^2=C Q^2 \\
\Rightarrow & (\alpha-3)^2+\alpha^2=(\alpha+2)^2+(\alpha-3)^2 \\
\therefore \quad & (\alpha+2)^2=\alpha^2 \Rightarrow 4 \alpha+4=0 \Rightarrow \alpha=-1 \\
\therefore \quad & \text { Centre }(-1,0) \Rightarrow \text { Radius }=\sqrt{4^2+1^2}=\sqrt{17}
\end{aligned}
$$
$\therefore$ Parametric equation is
$$
x=-1+\sqrt{17} \cos \theta \text { and } y=\sqrt{17} \sin \theta \text {. }
$$
$\therefore \quad$ Centre: $C(\alpha, \alpha+1)$
Let $P(3,1)$ and $Q(-2,4)$ are 2 points
$$
\begin{aligned}
& \therefore \quad C P=C Q \Rightarrow C P^2=C Q^2 \\
\Rightarrow & (\alpha-3)^2+\alpha^2=(\alpha+2)^2+(\alpha-3)^2 \\
\therefore \quad & (\alpha+2)^2=\alpha^2 \Rightarrow 4 \alpha+4=0 \Rightarrow \alpha=-1 \\
\therefore \quad & \text { Centre }(-1,0) \Rightarrow \text { Radius }=\sqrt{4^2+1^2}=\sqrt{17}
\end{aligned}
$$
$\therefore$ Parametric equation is
$$
x=-1+\sqrt{17} \cos \theta \text { and } y=\sqrt{17} \sin \theta \text {. }
$$
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