In a triangle PQR, the co-ordinates of the points P and Q are (-2,4) and (4,-2) respectively. If the

Q: Wait, why do we equate $CP^2 = CQ^2$ instead of just $CP = CQ$? Won't that change the answer?

That is a brilliant observation! Squaring both sides is a strategic maneuver to eliminate the square roots from the distance formula. Since distances are always non-negative, $CP = CQ$ is mathematically equivalent to $CP^2 = CQ^2$. It saves you from dealing with messy radicals and potential sign errors, making your algebra much cleaner.

Q: Do I really need to find the equation of the perpendicular bisector of $PQ$ explicitly? Can't I just use the slope of $PQ$?

You absolutely can! You could find the midpoint of $PQ$, calculate the negative reciprocal of the slope of $PQ$, and use the point-slope form. However, using the equidistance property ($CP^2 = CQ^2$) is often faster and less prone to calculation errors. Both methods are valid, but the equidistance method is a powerful tool in your JEE arsenal.

Q: Why does the circumcenter have to lie on the perpendicular bisector of $PR$? Is that always true?

Yes, it is always true! By definition, the circumcenter is the intersection of the perpendicular bisectors of the sides of the triangle. Since $PR$ is a side of $\triangle PQR$, the circumcenter must lie on its perpendicular bisector. This is a geometric certainty, not a coincidence.

Question

In a triangle PQR, the co-ordinates of the points P and Q are (-2,4) and (4,-2) respectively. If the equation of the perpendicular bisector of PR is $2x - y + 2 = 0$, then the centre of the circumcircle of the $	riangle PQR$ is :

Options:

(A) (-1,0)

(B) (-2,-2)

(C) (0,2)

(D) (1,4)

Sigma Percentile · Accepted Answer

### The Geometry of the Circumcenter: A Journey of Symmetry

Welcome, future engineer! Today, we are going to unravel the mystery of the circumcenter of a triangle.

Imagine you are standing on a coordinate plane, looking at a triangle $	riangle PQR$ with vertices $P(-2, 4)$ and $Q(4, -2)$. We are given a secret key: the perpendicular bisector of the side $PR$ is $2x - y + 2 = 0$. Our mission is to find the circumcenter $C(h, k)$.

### Phase 1: Defining the Battlefield

What is a circumcenter? It is the heart of the triangle, the point where the perpendicular bisectors of all three sides meet.

More importantly, it is the center of the circle that passes through all three vertices. This implies a beautiful property: the distance from the circumcenter $C$ to any vertex is the radius of the circumcircle.

Thus, $CP = CQ = CR$. This equidistance is our most powerful weapon.

### Phase 2: The First Constraint

We are told that the perpendicular bisector of $PR$ is $2x - y + 2 = 0$. Since the circumcenter $C(h, k)$ must lie on this bisector, it must satisfy the equation.

By substituting $h$ and $k$ into the line equation, we get our first constraint:

$$2h - k + 2 = 0 \quad \dots(1)$$

This is our anchor. We have one equation with two variables, so we need one more to solve the system.

### Phase 3: The Hidden Symmetry

How do we find the second constraint? We look at the side $PQ$. We know that $C$ is equidistant from $P$ and $Q$, so $CP = CQ$.

To avoid the terror of square roots, we square both sides:

$$CP^2 = CQ^2$$

Using the distance formula, we write:

$$(h - (-2))^2 + (k - 4)^2 = (h - 4)^2 + (k - (-2))^2$$

Let's expand this carefully. The left side becomes $h^2 + 4h + 4 + k^2 - 8k + 16$, and the right side becomes $h^2 - 8h + 16 + k^2 + 4k + 4$.

Look at the beauty of the cancellation! The $h^2$, $k^2$, and the constant $20$ appear on both sides. They vanish, leaving us with:

$$4h - 8k = -8h + 4k$$

### Phase 4: The Final Intersection

Rearranging this, we get $12h = 12k$, which simplifies to the elegant relation:

$$h = k \quad \dots(2)$$

Now, we have a system of two simple linear equations. Substituting $k = h$ into our first equation ($2h - k + 2 = 0$), we get $2h - h + 2 = 0$, which simplifies to $h + 2 = 0$.

Thus, $h = -2$. Since $h = k$, we immediately find $k = -2$.

The circumcenter is **$(-2, -2)$**.

### Conclusion

We have successfully navigated the geometry. By leveraging the definition of the circumcenter and the power of the equidistance property, we turned a complex problem into a simple system of equations.

Remember, in JEE Advanced, the math is rarely the obstacle; it is the clarity of your geometric intuition. Keep practicing, and keep visualizing!

Animated Solution for Mathematics - Straight Lines: In a triangle PQR, the co-ordinates of the points P and Q are (-2,4) and (4,-2) respectively. If the equation of the perpendicular bisector of PR is $2 x - y + 2 = 0$ , then the centre of the circumcircle of the $△ P QR$ is :

Select Answer:

Visualized Solution

Understanding the Circumcenter

Plotting Vertices $P$ and $Q$

The First Constraint: Bisector of $P R$

The Equidistance Property

Applying the Distance Formula

Expanding the Squares

Simplifying the Equation

Finding the Second Constraint

Solving the System of Equations

Finding the Coordinates of $C$

Final Conclusion

The Sigma Insight: Centroid, Incenter, Orthocenter, and Circumcenter

The Geometry of the Circumcenter

Phase 1

Phase 2

Phase 3

Phase 4

Conclusion

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