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If the straight line $x \cos \alpha+y \sin \alpha=P$ intersects the circle $x^2+y^2=a^2$ at $A$ and $B$, then the equation of the circle with diameter $\overline{A B}$ is
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Verified Answer
The correct answer is:
$x^2+y^2-2 P x \cos \alpha-2 P y \sin \alpha+2 P^2-a^2=0$
Let the circle be
$$
(\text { as } s+\lambda L=0)
$$
Its centre is $\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)$
$\because$ Centre lies on $x \cos \alpha+y \sin \alpha=P$
$$
\begin{aligned}
& \therefore \frac{(-\lambda \cos \alpha)}{2} \cos \alpha+\left(-\frac{\lambda \sin \alpha}{2}\right) \sin \alpha=P \\
& \Rightarrow \lambda=-2 P
\end{aligned}
$$
From Eq. (i),
$$
x^2+y^2-2 P x \cos \alpha-2 P y \sin \alpha+2 P^2-a^2=0
$$
$$
(\text { as } s+\lambda L=0)
$$
Its centre is $\left(-\frac{\lambda \cos \alpha}{2},-\frac{\lambda \sin \alpha}{2}\right)$
$\because$ Centre lies on $x \cos \alpha+y \sin \alpha=P$
$$
\begin{aligned}
& \therefore \frac{(-\lambda \cos \alpha)}{2} \cos \alpha+\left(-\frac{\lambda \sin \alpha}{2}\right) \sin \alpha=P \\
& \Rightarrow \lambda=-2 P
\end{aligned}
$$
From Eq. (i),
$$
x^2+y^2-2 P x \cos \alpha-2 P y \sin \alpha+2 P^2-a^2=0
$$
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