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The straight line $\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha=\mathrm{p}$ cuts the circle $x^2+y^2-a^2=0$ at $A$ and $B$. Then the equation of circle having $\mathrm{AB}$ as diameter is
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The correct answer is:
$x^2+y^2-a^2-2 p(x \cos \alpha+y \sin \alpha-p)=0$
Since Centre of the circle having diameter as AB.
Now foot of perpendicular from $(0,0)$ to the line $\mathrm{x} \cos$ $\alpha+\mathrm{y} \sin \alpha=\mathrm{p}$ is
$\begin{aligned} & \frac{\mathrm{x}-0}{\cos \alpha}=\frac{\mathrm{y}-0}{\sin \alpha}=\frac{(-\mathrm{P})}{1} \\ & \Rightarrow \mathrm{x}=\mathrm{p} \cos \alpha, \mathrm{y}=\mathrm{P} \sin \alpha\end{aligned}$
how equation of circle having centre $(\mathrm{P} \cos \alpha, \mathrm{P} \sin$ $\alpha)$ and radius $(r)=\sqrt{a^2-p^2}$ is
$(\mathrm{x}-\mathrm{p} \cos \alpha)^2+(\mathrm{y}-\mathrm{p} \sin \alpha)^2=\left(\sqrt{\mathrm{a}^2-\mathrm{p}}\right)^2$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2-2 \mathrm{p}(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})=0$
Now foot of perpendicular from $(0,0)$ to the line $\mathrm{x} \cos$ $\alpha+\mathrm{y} \sin \alpha=\mathrm{p}$ is
$\begin{aligned} & \frac{\mathrm{x}-0}{\cos \alpha}=\frac{\mathrm{y}-0}{\sin \alpha}=\frac{(-\mathrm{P})}{1} \\ & \Rightarrow \mathrm{x}=\mathrm{p} \cos \alpha, \mathrm{y}=\mathrm{P} \sin \alpha\end{aligned}$
how equation of circle having centre $(\mathrm{P} \cos \alpha, \mathrm{P} \sin$ $\alpha)$ and radius $(r)=\sqrt{a^2-p^2}$ is
$(\mathrm{x}-\mathrm{p} \cos \alpha)^2+(\mathrm{y}-\mathrm{p} \sin \alpha)^2=\left(\sqrt{\mathrm{a}^2-\mathrm{p}}\right)^2$
$\Rightarrow \mathrm{x}^2+\mathrm{y}^2-\mathrm{a}^2-2 \mathrm{p}(\mathrm{x} \cos \alpha+\mathrm{y} \sin \alpha-\mathrm{p})=0$
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