Let the abscissae of the two points P and Q on a circle be the roots of x² - 4x - 6 = 0 and the ordi

Q: Wait, why can't I just solve for the roots $x_1, x_2$ and $y_1, y_2$ using the quadratic formula and then find the center and radius?

You absolutely could, but that is the long, treacherous path! The roots of $x^2 - 4x - 6 = 0$ are irrational, and calculating with them will lead to messy algebra and a high probability of arithmetic errors. In JEE Advanced, the goal is to identify the elegant, structural shortcut. By using the diametric form, we bypass the need to know the individual coordinates entirely.

Q: Is the formula $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$ always valid for any two points on a circle?

No, that is a crucial distinction! This formula is valid ONLY when the line segment joining $(x_1, y_1)$ and $(x_2, y_2)$ is a diameter of the circle. This is because the angle subtended by a diameter at any point on the circle is $90^\circ$. If the segment were just a chord, this equation would not represent the circle.

Q: How do I compare my simplified equation to the general form $x^2 + y^2 + 2ax + 2by + c = 0$?

Once you have your equation in the form $x^2 + y^2 + Dx + Ey + F = 0$, you simply map the coefficients. Here, $2a$ corresponds to the coefficient of $x$, $2b$ to the coefficient of $y$, and $c$ to the constant term. Just be careful to include the signs when equating them!

Question

Let the abscissae of the two points $P$ and $Q$ on a circle be the roots of $x^2 - 4x - 6 = 0$ and the ordinates of $P$ and $Q$ be the roots of $y^2 + 2y - 7 = 0$. If $PQ$ is a diameter of the circle $x^2 + y^2 + 2ax + 2by + c = 0$, then the value of $(a+b-c)$ is

Options:

(A) 12

(B) 13

(C) 14

(D) 16

Sigma Percentile · Accepted Answer

### The Geometry of the Diameter: A Masterclass in Elegance

Welcome, fellow explorers of the coordinate plane. Today, we are going to dismantle a classic JEE Advanced problem that, at first glance, might tempt you to reach for the quadratic formula.

In the world of competitive mathematics, the most powerful tool is not always the most complex calculation—it is the deepest geometric insight.

### The Hidden Structure

We are given two points, $P(x_1, y_1)$ and $Q(x_2, y_2)$, which lie on a circle. We are told that $PQ$ is the diameter.

We are also given two quadratic equations: $x^2 - 4x - 6 = 0$, whose roots are the abscissae ($x_1, x_2$), and $y^2 + 2y - 7 = 0$, whose roots are the ordinates ($y_1, y_2$).

Most students will immediately try to solve for $x_1, x_2, y_1,$ and $y_2$. I urge you to resist this urge.

The problem does not ask for the coordinates of $P$ and $Q$; it asks for the value of $(a+b-c)$ derived from the circle's equation. This is a massive hint that the coordinates themselves are irrelevant, and we only need the equations that define them.

### The Diametric Form: The Shortcut

Recall the fundamental property of a circle: any diameter subtends a right angle at any point on the circumference. If we take any point $(x, y)$ on the circle, the slope of $PX$ multiplied by the slope of $QX$ must equal $-1$.

This geometric truth leads us to the beautiful diametric form of a circle's equation:

$$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$$

Look at this equation; it is a masterpiece of symmetry. We know that $x_1$ and $x_2$ are the roots of $x^2 - 4x - 6 = 0$.

By the property of quadratic equations, we know that $(x - x_1)(x - x_2)$ is identically equal to $x^2 - 4x - 6$. Similarly, $(y - y_1)(y - y_2)$ is identically equal to $y^2 + 2y - 7$.

### The Synthesis

Now, watch the magic happen. We substitute these quadratic expressions directly into our diametric form:

$$(x^2 - 4x - 6) + (y^2 + 2y - 7) = 0$$

There is no need for expansion or messy root calculations. We simply group the terms to reveal the circle's equation:

$$x^2 + y^2 - 4x + 2y - 13 = 0$$

This is the equation of our circle. It is clean, it is precise, and it was achieved without ever knowing exactly where $P$ and $Q$ sit on the plane.

### The Final Comparison

The problem asks us to compare this with the general form $x^2 + y^2 + 2ax + 2by + c = 0$. By aligning the coefficients, we get:

$$2a = -4 \Rightarrow a = -2$$

$$2b = 2 \Rightarrow b = 1$$

$$c = -13$$

Finally, we calculate the target expression $(a + b - c)$:

$$a + b - c = (-2) + (1) - (-13) = -2 + 1 + 13 = 12$$

### The Takeaway

We have arrived at our answer: **$12$**. Notice how the complexity of the problem vanished the moment we stopped treating it as an algebra problem and started treating it as a geometry problem.

In JEE Advanced, always look for the structure. When you see roots of quadratics defining points on a circle, don't solve for the points—solve for the circle. Keep this perspective, and you will find that even the most intimidating problems become a series of elegant, logical steps.

Select Answer:

Visualized Solution

Visualizing the Points $P$ and $Q$

Defining the Abscissae

Defining the Ordinates

The Diametric Form of a Circle

Substituting the Quadratic Expressions

Simplifying the Circle Equation

Comparing with the General Form

Finding the Values of $a, b,$ and $c$

Final Calculation of $a + b - c$

The Sigma Insight: Standard and General Equation of a Circle

The Geometry of the Diameter

The Hidden Structure

The Diametric Form

The Synthesis

The Final Comparison

The Takeaway

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