Let P and Q be two distinct points on a circl | vedclass

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(A) The center of the circle is $C(2,3)$ and it passes through the origin $O(0,0)$. The radius $r$ is the distance $OC = \sqrt{2^2 + 3^2} = \sqrt{13}$

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$\{(-1,5), (5,1)\}$

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(A) The center of the circle is $C(2,3)$ and it passes through the origin $O(0,0)$. The radius $r$ is the distance $OC = \sqrt{2^2 + 3^2} = \sqrt{13}$.The slope of $OC$ is $m_{OC} = \frac{3-0}{2-0} = \frac{3}{2}$.Since $CP \perp OC$ and $CQ \perp OC$,the line $PQ$ is perpendicular to $OC$. The slope of line $PQ$ is $m = -\frac{1}{m_{OC}} = -\frac{2}{3}$.Using the parametric form of a line passing through $C(2,3)$ with slope $m = -\frac{2}{3}$,we have $\cos 	heta = \frac{3}{\sqrt{13}}$ and $\sin 	heta = -\frac{2}{\sqrt{13}}$ (or vice versa).The coordinates of $P$ and $Q$ are $(x, y) = (2 \pm r \cos 	heta, 3 \pm r \sin 	heta)$.Substituting $r = \sqrt{13}$,$\cos 	heta = \frac{3}{\sqrt{13}}$,and $\sin 	heta = -\frac{2}{\sqrt{13}}$:$x = 2 \pm \sqrt{13} \left(\frac{3}{\sqrt{13}}ight) = 2 \pm 3 \implies x = 5 	ext{ or } -1$.$y = 3 \pm \sqrt{13} \left(-\frac{2}{\sqrt{13}}ight) = 3 \mp 2 \implies y = 1 	ext{ or } 5$.Thus,the points are $(5, 1)$ and $(-1, 5)$.

Let $P$ and $Q$ be two distinct points on a circle which has center at $C(2,3)$ and which passes through the origin $O(0,0)$. If $OC$ is perpendicular to both the line segments $CP$ and $CQ$,then the set $\{P, Q\}$ is equal to:

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