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

A Journey to the Center

Welcome, future engineers! Today, we are going to unravel a beautiful problem that sits at the intersection of coordinate geometry and pure intuition. We are dealing with a circle, a center $C(2,3)$, and a path that takes us through the origin $O(0,0)$.

At first glance, this might seem like a standard coordinate geometry exercise, but there is a hidden elegance here that we are going to uncover together.

The Foundation

Every great journey begins with a solid foundation. We know our circle has its center at $C(2,3)$ and passes through the origin $O(0,0)$.

The distance from the center to any point on the circumference is the radius, $r$. So, the distance $OC$ is our radius. Using the distance formula, we calculate:

There we have it! Our circle is defined by the equation $(x-2{)}^2+(y-3{)}^2=13$.

But hold on—don't rush into expanding that equation just yet. Geometry often rewards those who pause to visualize.

The Geometric Insight

The problem gives us a crucial clue: $OC\perp CP$ and $OC\perp CQ$. Imagine standing at the center $C$. You have a line segment $OC$ pointing toward the origin.

Now, you have two other segments, $CP$ and $CQ$, both of which are perpendicular to $OC$ at the point $C$. If two lines are perpendicular to the same line at the same point, they must be collinear.

This means $P,C,$ and $Q$ form a single straight line. Since $P$ and $Q$ are on the circle and the line $PQ$ passes through the center $C$, $PQ$ is not just any chord—it is a diameter!

The Power of Parametric Equations

Now, we need to find the coordinates of $P$ and $Q$. We know the slope of $OC$ is $m_1=\frac{3-0}{2-0}=\frac{3}{2}$.

Since $PQ$ is perpendicular to $OC$, the slope of our diameter $PQ$ must be the negative reciprocal: $m_2=-\frac{1}{m_1}=-\frac{2}{3}$.

Instead of solving a messy system of equations, let's use the parametric form of a line. For any point on a circle with center $(x_c,y_c)$ and radius $r$, the coordinates can be expressed as:

Here, $\tan \theta =-\frac{2}{3}$. Since $\tan \theta =\frac{\sin \theta }{\cos \theta }=-\frac{2}{3}$, we can represent $\sin \theta$ and $\cos \theta$ using a right triangle where the opposite side is $2$ and the adjacent side is $3$.

Thus, the hypotenuse is $\sqrt{2^2+3^2}=\sqrt{13}$. This gives us $\cos \theta =\frac{3}{\sqrt{13}}$ and $\sin \theta =-\frac{2}{\sqrt{13}}$ (or vice versa for the signs).

The Final Victory

Now, let's plug these values into our parametric equations:

Calculating these gives us our two points:

For the plus sign: $x=2+3=5$ and $y=3-2=1$. So, $(5,1)$.

For the minus sign: $x=2-3=-1$ and $y=3+2=5$. So, $(-1,5)$.

And there it is! The set of points $\{P,Q\}$ is $\{(5,1),(-1,5)\}$.

By using the geometric property of the diameter and the elegance of parametric equations, we bypassed the tedious algebra and arrived at the solution with clarity and confidence. Keep practicing this, and you will find that geometry is not just about formulas—it is about seeing the hidden structure of the world.