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Step 1: Determine the radius of the circle
The circle has center at $C(2,3)$ and passes through the origin $O(0,0)$. The radius $r$ is the distance from $C$ to $O$, which is
$r = \sqrt{(2-0)^2 + (3-0)^2} = \sqrt{4 + 9} = \sqrt{13}.$
Step 2: Impose the condition for perpendicularity
We are given that $OC$ is perpendicular to $CP$ (and also to $CQ$). Let $P = (x,y)$ be a point on the circle. Then:
Vector $OC$ is $(-2, -3)$ since $C(2,3)$ and $O(0,0)$.
Vector $CP$ is $(x - 2,\; y - 3)$.
For perpendicularity, their dot product must be zero:
$OC \cdot CP = (-2)(x - 2) + (-3)(y - 3) = 0.$
So,
$-2x + 4 - 3y + 9 = 0 \quad \Longrightarrow \quad -2x - 3y + 13 = 0 \quad \Longrightarrow \quad 2x + 3y = 13.$
Step 3: Impose the condition for $P$ to lie on the circle
Since $P$ is on the circle with center $C(2,3)$ and radius $\sqrt{13}$, it must satisfy:
$(x - 2)^2 + (y - 3)^2 = 13.$
Also from the perpendicularity condition, $y = \frac{13 - 2x}{3}.$ We substitute this into the circle equation to find $x$ and $y$.
Step 4: Solve the system of equations
Linear equation: $y = \frac{13 - 2x}{3}.$
Circle equation: $(x - 2)^2 + (y - 3)^2 = 13.$
By substituting $y$ into the circle equation, we solve for $x$ and then for $y$. This yields two solutions, which correspond to the points $P$ and $Q$.
Upon solving, we obtain
$P = (-1,5) \quad \text{and} \quad Q = (5,1).$
Step 5: Verify that both points satisfy all conditions
1. Both points lie on the circle with radius $\sqrt{13}$ centered at $(2,3)$.
2. The vector $OC$ is perpendicular to $CP$ and $CQ$ for these points.
3. Hence, the required points are $(-1,5)$ and $(5,1)$.
Final Answer
Therefore, the set $\{P, Q\}$ is
$\{(-1,5), (5,1)\}.$
