Animated Solution for Mathematics - Circles: The triangle $PQR$ is inscribed in the circle $x^2+y^2=25$. If $Q$ and $R$ have co-ordinates $(3,4)$ and $(-4,3)$ respectively, then $\angle QPR$ is equal to

Welcome, future engineer. Today, we are not just solving a problem; we are embarking on a journey through the elegant symmetry of coordinate geometry.

When you look at a problem like this, involving a triangle inscribed in a circle, I want you to stop seeing just numbers and variables. I want you to see a story of relationships.

We have a circle, defined by the equation $x^2+y^2=25$. This is the bedrock of our problem. It tells us immediately that we are dealing with a perfect circle centered at the origin $O(0,0)$ with a radius $r=5$.

Imagine standing at the center of this circle. You look out at two points, $Q(3,4)$ and $R(-4,3)$, sitting on the circumference. You are asked to find the angle $\angle QPR$, where $P$ is some arbitrary point on the major arc.

At first glance, you might feel a sense of unease. Where is $P$? Does its position change the answer?

This is the first trap of the JEE. The beauty of geometry is that some things are invariant—they do not change, no matter how much you move the pieces. Let us uncover why.

To understand the angle at the circumference, we must first understand the angle at the heart of the circle. Let us connect the center $O$ to our known points $Q$ and $R$.

By doing this, we create two radii, $OQ$ and $OR$. This forms the central angle $\angle QOR$.

Why do we do this? Because there is a fundamental theorem in geometry: the angle subtended by an arc at the center is exactly twice the angle subtended by it at any point on the remaining part of the circle. If we find $\angle QOR$, we have effectively found $\angle QPR$.

Now, how do we find the angle between two lines, $OQ$ and $OR$? We use the language of slopes. The slope $m$ of a line passing through $(x_1,y_1)$ and $(x_2,y_2)$ is given by $m=\frac{y_2-y_1}{x_2-x_1}$.

Let us apply this to our radii. For the radius $OQ$, connecting $(0,0)$ and $(3,4)$, the slope $m_{OQ}$ is:

Now, let us turn our attention to the radius $OR$, connecting $(0,0)$ and $(-4,3)$. The slope $m_{OR}$ is:

Look closely at these two values: $\frac{4}{3}$ and $-\frac{3}{4}$. Do you see the magic happening? When you multiply these two slopes together, something extraordinary occurs:

In the world of coordinate geometry, the condition $m_1\cdot m_2=-1$ is the hallmark of perpendicularity. It tells us, with absolute mathematical certainty, that the radius $OQ$ is perpendicular to the radius $OR$.

This means the central angle $\angle QOR$ is exactly $90^{\circ }$. This is the moment where the problem collapses into simplicity.

We have found the central angle. Now, we invoke the Inscribed Angle Theorem. We know that:

Rearranging this, we find the target angle:

We have arrived at $45^{\circ }$. But look at your options. They are expressed in radians.

In the JEE, precision is not just about the value; it is about the format. We convert degrees to radians by multiplying by $\frac{\pi }{180^{\circ }}$:

And there it is. The answer is $\frac{\pi }{4}$.

Think about what we just did. We didn't need to know the coordinates of $P$. We didn't need to perform complex trigonometry. We simply used the properties of the circle to reveal a hidden right angle.

This is the essence of physics and mathematics—finding the underlying structure that simplifies the chaos. Keep this mindset, and you will conquer any problem the exam throws at you.